Group Theory G13GTH. Chris Wuthrich. 1 Definitions and examples Examples Direct and semi-direct products Small groups...

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1 Group Theory G13GTH Chris Wuthrich Contents 1 Definitions and examples Examples Direct and semi-direct products Small groups Group actions Action on cosets Action by conjugation Linear representations A second look at semi-direct products Metacyclic groups The symmetric group The sign of permutations The alternating group Sylow s theorems Applications Finitely generated abelian groups Free abelian groups Smith normal form The fundamental theorem on finitely generated abelian groups Series Composition series Soluble groups Nilpotent groups Primitive actions 35 8 Projective linear group Transvections

2 9 Finite rotation groups Two-dimensional case Three-dimensional examples Infinite families Coxeter s theorem

3 1 Definitions and examples A group is a non-empty set with a binary operation such that for all g, h, k G gh G; (gh)k = g(hk); there is a neutral element 1 G such that 1 g = g 1 = g; there is a g 1 G such that g g 1 = g 1 g = 1. If the group is abelian, i.e. when gh = hg for all g and h in G, we sometimes write the law as + and the neutral element as 0. This module on group theory is the continuation of the study of groups started in G11MSS and G12ALN. So many notions and definitions will be assumed as prerequisites. See the additional hand-out which recalls the definitions and basic properties of the following notions: Subgroup, coset, index, normal subgroup, quotient group, conjugate of elements and groups, homomorphism, isomorphism, kernel, image, generating set and subgroup generated by a set, etc. Notations: We will write H < G for a subgroup different from G; if H is allowed to be equal to G, we write H G. When writing N G we mean that N < G is a normal subgroup of G. The number of elements of a set X is denoted by #X. However if G is a group then we also write G = #G and call it the order of G. By cosets of H G, we will always mean left cosets gh with g G. Similar the conjugation of g on h is ghg 1. Also all our actions (see section 2) are on the left. You will find that some books do everything on the right instead, for instance their conjugation is g 1 hg and so on. The integers modulo n, will be denoted by Z / nz. If p is a prime number, then Z / pz is sometimes also denoted by F p when we think of it as a field. 1.1 Examples The symmetric groups S X and S n If X is a set, then the set S X of all bijections X X is a group under composition. If X = {1, 2,..., n}, then we write S n and we call it the symmetric group of degree n. See section 3 for much more on this group and it interesting subgroup A n. The cyclic group C n For any integer n 1, the cyclic group of order n is the set {1, g, g 2, g 3,..., g n 1 } with the operation { g i g j g i+j if i + j < n and = g i+j n otherwise. The element g is called a generator. 3

4 Integers modulo n For any integer n 1, the set of integers modulo n, denoted by Z / nz, is the quotient group of Z by its subgroup nz. Its law is written + and the neutral element is 0. It is isomorphic to the cyclic group C n. The units R For any ring R, the set of units R is a group under multiplication. Here a element r of R is a unit (or invertible element) if there is a s R such that rs = 1. If R is a field, like when R = F p is the field of p elements for some prime p, then R = R \ {0}. You might see that F p is isomorphic to a cyclic group of order p 1 in G13FNT or G13NGA. The general linear group GL n (R) Let R be a ring and let n 1. Then the set of all n n matrices with coefficients in R such that their determinant is a unit in R is a group under matrix multiplication, called the general linear group GL n (R). If R is a field, this is all matrices with non-zero determinants; if R = Z this is the set of all matrices with determinant ±1. The subgroup of all matrices with determinant 1 is called the special linear group SL n (R). The dihedral group D n For any n > 1, the dihedral group D n is the set of all isometries, i.e. distance preserving maps, of the plane that map a regular n-gon to itself. The neutral element is the identity map and the operation is composition. The group D n is composed of n rotations (including id) and n reflections. So D n = 2n. (In some books this group is denoted by D 2n.) Isometry groups If X is any subset of R n for some n 1, we can look at the set of all isometries Isom(X) that preserves X. The subset of all orientation-preserving isometries is a normal subgroup. The automorphism group Let G be a group. Then the set of all isomorphisms G G is called the automorphism group Aut(G) of G. The identity id: G G is the neutral element. 1.2 Direct and semi-direct products Let G and H be two groups. The direct product G H is the set { (g, h) g G, h H } with the operation (g, h)(g, h) = (gg, hh ) for all g, g G and h, h H. For instance, C n C m = Cnm if n and m are coprime. If G and H are finite then G H = G H. Theorem 1.1. Let H and K be two subgroups of a group G. Suppose that (a). hk = kh for all h H and k K and (b). each element g G can be written uniquely as g = hk for some h H and k K. 4

5 Then G is isomorphic to H K. Proof. Let g G. By (b) there is a h H and a k K such that g = hk. Define a map ϕ(g) = (h, k) from G to (H, K). By assumption (b) this is a well-defined bijection. If g = hk and g = h k for h, h H and k, k K, then gg = hkh k = hh kk by (a) and so ϕ(gg ) = ϕ(g) ϕ(g ) shows that ϕ is a homomorphism. Generalisations of this theorem to multiple products H 1 H 2 H k are immediate. The condition (b) can also be rephrased by asking that G = HK and H K = {1}; see the lemma 1.3 below. Proposition 1.2. Let {1} = G be a finite group such that g 2 = 1 for all g in G. Then G = C 2 C 2 C 2. Proof by induction on G. If G = 2, then G = C 2 and we are done. Suppose G > 2. For any g, h G, we have (gh) 2 = 1 and therefore gh = ghgg = ghghhg = 1hg = hg shows that G is abelian. Let {g 1, g 2,..., g k } be a minimal set of generators of G. Set g = g 1 and H = g 2, g 3,..., g k. By minimality g can not belong to H. Now every element of G can be expressed as words in g i and by grouping together those in H, remembering that G is abelian, we see that every element of G can be written as hg n with n {0, 1} and h H. This representation is unique since hg = h g implies h = h for h, h H and hg = h implies that g H which is impossible. Hence by the previous theorem, G = H g = H C 2. By induction H is a product of a finite number of copies of C 2 and hence so is G. Let G be a group. Suppose N G and H < G such that every element in G can be uniquely written as g = nh with n N and h H. Then G is called a semi-direct product of N and H, written as N H. Lemma 1.3. Let N G and H < G. Then G = N H if and only if G = NH and N H = {1}. The proof is an exercise on the problem sheet. We will see later in section 2.4 how given two groups H and N and some extra data, we can form a group that is a semi-direct product of N and H. Proposition 1.4. Let n 3. The dihedral group D n is the semi-direct product N H with N = C n being the subgroup of rotations and H = C 2 is generated by any choice of a reflection g D n. Proof. Since N has index 2, it is normal. If h is any reflection, then there is a rotation t that brings the axis of the reflection h to the axis of the reflection g. The rotation t may not belong to D n when n is even, but t 2 N. We claim that h = t 2 g. The picture in figure 1 illustrates this. Take any point x, reflect it with g to get g(x) and then rotate it twice with t. The three points x, g(x), and t 2 g(x) have all the same distance from the origin. Moreover it is easy to see that the angle between x and the axis of relection of h and the angle between t 2 g(x) and the axis of reflection of h are equal. Hence t 2 g(x) = h(x). It follows that D n = NH. Finally N H = {1} shows by lemma 1.3 that D n = N H. 5

6 Figure 1: Bringing one reflection to the other 1.3 Small groups Recall that if n is prime number then there is only one isomorphism class of groups of order n, namely C n. Groups of order 4 There are exactly two groups of order 4, namely C 4 and C 2 C 2, and both are abelian. To see this, note that if a group G of order four has an element of order 4, then it is C 4, otherwise proposition 1.2 applies. Groups of order 6 There are two non-isomorphic groups of order 6, namely C 6 and S 3 = D 3. Let us prove this: Let G be a group of order 6. Elements have order 1, 2, 3 or 6 in G by Lagrange s theorem. Not all elements have order 1 and 2, because of proposition 1.2 and the fact that 6 is not a power of two. Even if there is an element of order 6, there must be an element g of order 3. It generates a subgroup of index 2, hence it is normal. Let h be an element not in this subgroup. Then h 2 must lie in g, but if h 2 is of order 3 then h is of order 6 and G = C 6. Otherwise h 2 = 1 and we have found that G = g h is D 3. Groups of order 8 There are 5 groups of order 8. Three of them, namely C 8, C 4 C 2 and C 2 C 2 C 2 are abelian. Then there is the non-abelian D 4 and a fifth non-abelian group Q 8. 6

7 The quaternion group Q 8 can be generated by the two elements g = ( ) i 0 0 i and h = ( ) as a subgroup of GL2 (C) where i 2 = 1. In the problem sheet you will show that every subgroup of Q 8 is normal and that it is not a semi-direct product of any its subgroups. It can also be described as the unit group of a non-commutative ring H called the Hurwitz quaternions. Groups of order 9 All groups of order 9 are abelian, either isomorphic to C 9 or C 3 C 3. Groups of order 10 There are two groups of order 10, namely C 10 and D 5. Groups of order 12 There are 5 groups of order 12. First the abelian groups C 12 and C 2 C 6. Then the nonabelian groups D 6 and A 4 as well as a non-abelian group which is a semi-direct product C 3 C 4 which we will construct in section 2.4. List of small groups The number of groups (up to isomorphism of course) of a given order n is listed as the first sequence in the online encyclopedia of integer sequences. Here are the first few values: n # n #

8 2 Group actions Most groups that we encounter in mathematics arise as groups of transformations on some set X. For instance the dihedral group D n is a group of transformations of the plane. Galois groups, as you may see in G13NGA, are transformations of solutions of polynomial equations, etc. We now define the notion of a group G acting on a set X. To give such an action is to view each element g G as a map T g : X X such that T h ( Tg (x) ) = T hg (x) and T 1 (x) = x for all g, h G and x X. (2.1) It is more convenient to write T g (x) as g x. The group action is then a map G X X such that h (g x) = (hg) x and 1 x = x for all g, h G and x X. (2.2) We defined a left action here, in some books you will find right actions instead. Note that T g 1 T g = T 1 = id X is the identity on X and hence T g and T g 1 are inverse to each other. In other words T g is an element of S X for all g. The condition (2.1) can be rephrased yet again, by saying that ρ: G S X g T g is a group homomorphism. setting g x = ρ(g)(x). Conversely any such homomorphism ρ gives an action by The cyclic group G = C n acts on X = C. Choose a generator g of C n and define for each 0 k < n and z C the action by g k z = e 2πi k n z. The dihedral group G = D n acts on X = R 2 by rotations and reflections. Let X be a set. Any subgroup G S X acts naturally on X. Let k be a field and n 1. Any subgroup G GL n (k) acts naturally on k n as matrices multiply vectors. Lemma 2.1. Let G act on X. For x, y X, define the relation x y if and only if there is a g G such that g x = y. Then is an equivalence relation on X. The proof is an exercise. It follows that the action of G partitions the set X into equivalence classes for, which are called orbits of G on X, or G-orbits on X. We set for each x X Orb G (x) = { g x g G } X which is the orbit containing x. Often it is simply denoted by Gx. The stabiliser of x X is Sometimes it is denoted by G x. Stab G (x) = { g G g x = x } G. 8

9 Lemma 2.2. The stabiliser Stab G (x) is a subgroup of G. Proof. Since 1 x = x, we have 1 Stab G (x). Then for any g and h Stab G (x), we obtain (gh) x = g h x = g x = x and g 1 x = g 1 g x = (g 1 g) x = 1 x = x. Hence gh Stab G (x) and g 1 Stab G (x). As an example, we return to the action of D n on R 2. The point x = (0, 0) R 2 is never moved by an element of G. So Orb G (x) = {x} and Stab G (x) = G. The point x = (1, 0) is fixed only by the reflection through the x-axis. stabiliser is a subgroup of order 2. The orbit of x is a regular n-gon. So the A random point, like x = ( 5, π/24 ) R 2, is not fixed by any element of G other than 1. So the stabiliser is the trivial group and the orbit has 2n elements as in the picture. As a second example, we consider the cyclic group C 6 acting on X = {1, 2, 3, 4, 5, 6, 7}, where the generator g acts as the permutation (1 2 3)(4 5). The orbits are {1, 2, 3}, {4, 5}, {6}, and {7}. The stabiliser of any of the first three is the subgroup {1, g 3 } of order 2, the stabiliser of 4 and 5 is the subgroup {1, g 2, g 4 } of order 3 and the stabiliser of 6 or 7 is the full group G. Theorem 2.3 (Orbit-stabiliser theorem). Let G act on X and let x X. The number of elements in the orbit of x is equal to the index of the stabiliser of x, i.e. # Orb G (x) = G : Stab G (x). Another way to write the equality is Stab G (x) # Orb G (x) = G. Check this in all the examples above. Proof. Let Y be the set of left cosets of H = Stab G (x) in G. Hence #Y = G : H. Let g and g be in G. Then g x = g x g 1 g x = x g 1 g H g H = gh. (2.3) 9

10 Therefore, we can define the map Φ: Orb G (x) Y g x gh By the above it is well-defined and injective. As it is also surjective, it is a bijection. An action is called transitive on X if there is only one orbit. The action is called faithful if no two elements act the same way on X. It is not hard to see that the action is faithful if and only if ρ: G S X is injective. The action is called free if all stabilisers are trivial. Finally an action is called regular if it is transitive and free. See the problem sheets for more properties of these notions. 2.1 Action on cosets Let H G. We define an action of G on the set X of all left cosets of H by left multiplication: If g G and kh X then g (kh) := (gk)h is in X. Lemma 2.4. This action is transitive and the stabiliser of the coset kh is the conjugate khk 1 of H. Proof. Transitivity is clear. For g, k G we have g Stab G (kh) gkh = kh k 1 gk H g khk 1. Hence Stab G (kh) = khk 1. Theorem 2.5 (Cayley s theorem). Every finite group is isomorphic to a subgroup of a symmetric group S n. Proof. Take H = {1} in the above action. Now G acts transitively on itself by left multiplication and this action is clearly faithful. So there is an injective homomorphism ρ: G S G. The core of a subgroup H is defined to be the largest normal subgroup in G contained in H. Lemma 2.6. The core of H is equal to the intersection of all conjugates of H: Core(H) = ghg 1. In particular, it is the kernel of the homomorphism ρ: G S X for the action of G on the set X of left cosets of H. g G The proof is an exercise. We can refine Cayley s theorem. Theorem 2.7. Let G be a group and H a subgroup of finite index n. There is an injection of G/ Core(H) into the finite symmetric group S n. In particular every finite index subgroup H < G contains a normal subgroup N G of finite index. Moreover if N G is of finite index, then G/N injects into S n with n = G : N. 10

11 2.2 Action by conjugation We define the conjugation action of G on itself by g x = gxg 1. Check that this is indeed an action. It is clear from the definition that the orbit of x G is its conjugacy class {gxg 1 g G}. The stabiliser of x X = G is the centraliser of x: Stab G (x) = C G (x) = { g G gxg 1 = x } = { g G gx = xg }. Finally, the kernel of ρ: G S G is the centre Z(G) of G. Proposition 2.8. The number of elements in the conjugacy class of x is equal to the index of the centraliser of x. In particular it divides G. Theorem 2.9 (Class equation). Let G be a finite group. Pick in each of the k conjugacy classes an element g i for 1 i k. Then G = k G : C G (g i ). i=1 Sometimes it is convenient to separate this. The only conjugacy classes that contain only one element are those containing an element in the centre. So if we pick an element g i for 1 i l in all conjugacy classes with more than one element, then G = Z(G) + l G : CG (g i ). (2.4) Here is a variant of this action: G act on the set X of all its subgroups by conjugation g H = ghg 1. The orbit of a subgroup H is the set of all conjugates of H and the stabiliser is the normaliser of H i=1 N G (H) = { g G ghg 1 = H }. Lemma The normaliser N G (H) of H is the largest subgroup of G such that H N G (H). The orbit-stabiliser theorem implies that the number of conjugates of H is equal to the index of the normaliser N G (H) in G. This number is one if and only if H is normal in G. 2.3 Linear representations A linear representation of a group G over a field k is a vector space V over k with a linear action of G on it; this means that G acts on V such that g (λ v + w ) = λ g v + g w for all g G, λ k and v, w V. The dimension (or degree) of the representation is the dimension of V. The action is linear if and only if T g : V V is a linear transformation for every g G. Suppose dim V = n. If we choose a basis of V, we can write T g as an invertible n n 11

12 matrix with entries in k; in other words there is a group homomorphism ρ: G GL n (k). Conversely every such homomorphism gives rise to a linear action on k n. The example of C n acting on C discussed before is a 1-dimensional linear representation over C. Another example is Q 8 acting linearly on a 2-dimensional vector space over C by the matrices that defined the group. As an more detailed example, we will look at the action of D n acting on the 2-dimensional vector space R 2 again. Let σ be an element of order n, say the rotation by 2π n and let τ be the reflection along the x-axis. Then with respect to the standard basis, these correspond to the linear transformations T σ = ( cos( 2π n ) sin( 2π n ) sin( 2π n ) cos( 2π n ) ) T τ = ( ) Let k be any field and G a finite group. We take a G -dimensional vector space V over k and we label the basis elements by e g where g runs (in some order) through all the elements of G. We define a linear action on V by setting g e h = e gh for all g, h G. So this is just a permutation of the basis; then we extend the action linearly on all of V. This is called the regular representation and is often denoted by k[g]. Theorem Let k be a field. Every finite group is isomorphic to a subgroup of GL n (k) for some n = G. Proof. We only have to note that no element of g 1 acts trivially on k[g] and hence ρ: G GL n (k) is injective. For instance, we can look at G = C 3 and the field k = R. Let g be a generator of C 3. So V = R[C 3 ] has a basis {e 1, e g, e g 2}. The action by the three elements in G on V is given by T 1 = 0 1 0, T g = 1 0 0, and T g 2 = In fact, V contains two subspaces that are fixed by G. First W 1 = R(e 1 + e g + e g 2) is a one-dimensional R-vector space with the trivial (linear) action by C 3. Next W 2 = R(e 1 e g ) R(e g e g 2) is a two-dimensional R-vector space on which G = C 3 acts by rotations of angle 0, 2π 3, and 4π 3. The theory of linear representations only starts here. One can show that all such are direct sums of so-called irreducible representations. All these irreducible representations appear in C[G] and hence they can be totally classified. If one only looks at the traces χ(g) = Tr(T g ) of the transformation one get the so-called characters of G. A good introduction to this subject is given in the book Representations and characters of groups by James and Liebeck, QA171 JAM. 2.4 A second look at semi-direct products Let N and H be two groups. We say that H acts on N by automorphisms if there is an action of H on N such that each T h : N N is a group homomorphism. In other words if h (nn ) = (h n)(h n ) for all h H and n, n N. 12

13 Suppose H acts by automorphisms on a group N. Then we define a new group G: As a set it is just the set of all pairs (n, h) with n N and h H, but the multiplication is given by (n, h)(n, h ) = ( n (h n ), hh ). Lemma G is a group. Proof. It is one of the exercises to show that the operation is associative. The neutral element is obviously (1, 1) and the inverse of (n, h) is (h 1 n 1, h 1 ), because (n, h)(h 1 n 1, h 1 ) = ( n (h h 1 n 1 ), hh 1) = (nn 1, 1) = (1, 1). As a first example, we can take for any N and H with the trivial action: h n = n for all h H and n N. Then G is nothing but the direct product N H. If we take N any group and H = C 2 = h, we can define an action by h g = g 1 for all g N. Then this action is by automorphism if and only if N is abelian. The theorem below can be used to prove that the if N = C n then G = D n. Theorem (a). Let H act on N by automorphisms. Then the group G constructed above is the semi-direct product N H. (b). Conversely every G which is a semi-direct product of N G and H < G can be obtained in this way. Proof. (a): We view N = { (n, 1) n N } and H = { (1, h) h H } as a subgroups of ( G. Let g = (n, h) G and t = (n, 1) N. It is shown in a exercise that gtg 1 = n (h n ) n 1, 1 ) N. Hence N G. It is clear that G = NH and that N H = {1}. So G = N H by lemma 1.3. (b): Suppose G = N H. Because N G, we can define an action of H on N by conjugation h n = hnh 1 N in G. This action is by automorphisms because h (nn ) = hnn h 1 = hnh 1 hn h 1 = (h n)(h n ). Let G be the group constructed as above with this action. Then there is a bijection ϕ: G G sending g = nh to (n, h). Note that if h, h H and n, n N then the way to write nhn h as a product in NH is as ( nhn h 1)( hh ) with nhn h 1 N because N is normal. Finally ϕ ( nh ) ϕ ( n h ) = (n, h) (n, h ) = ( n (h n ), hh ) = ( nhn h 1, hh ) = ϕ ( nhn h ) shows that ϕ is a homomorphism. 2.5 Metacyclic groups Let p < q be two prime numbers with p dividing q 1. It is known from number theory (see G13FNT) that there is an element t Z / qz such that t p 1 (mod q) but t 1 (mod q). Define an action of Z / pz on Z / qz by (g + pz) (h + qz) = t g h + qz for g, h Z. 13

14 It is easy to see that this is an action by automorphism, i.e. (ḡ 1 + ḡ 2 ) ( h) = ḡ 1 (ḡ 2 h ) and ḡ ( h1 + h 2 ) = (ḡ h1 ) + (ḡ h2 ). The above construction yields a group Z / qz Z / pz of order pq. It is non-abelian as t 1 (mod q). Moreover the choice of t does not matter in that the two group constructed for two choices of t will be isomorphic. In general semi-direct products of the form Z / nz Z / mz are called a meta-cyclic group. They exist as soon as there is a non-trivial homomorphism from Z / mz to Aut (Z / nz ) = ( Z / nz ). 14

15 3 The symmetric group For any set X, let S X be the set of all bijections X X. If X = {1, 2,..., n}, we write S n for S X. They form a group under the composition g : X X and h: X X gives gh = g h: X X. Elements g in S n can be written as ( ) n. g(1) g(2) g(3) g(n) If 1 < k n, then a k-cycle g = (x 1 x 2... x k ) is the element such that g(x 1 ) = x 2, g(x 2 ) = x 3,..., g(x n ) = x 1. Two cycles (x 1 x 2... x k ) and (y 1 y 2... y l ) are disjoint if {x 1,... x k } {y 1,..., y l } =. Lemma 3.1. Every element can be written as a product of disjoint cycles. Up to reordering the factors, this is unique. Proof. Let g S n and consider the number of elements in {1,..., n} that are not fixed by g. If there are none then g = 1 is the empty product. If there are some elements moved by g, then we argue by induction. Pick a x X that is not fixed. Consider x, g(x), g(g(x)),.... Because X is finite, this sequence must start to repeat itself. Write h = (x 1 x 2... x k ) for this cycle. This cycle is non-trivial, i.e. k > 1 as otherwise our sequence would have reached an element fixed by g. Then g = h 1 g S n sends any x not appearing in this cycle to g(x) and fixes any x i in the cycle. Hence g moves less elements and can therefore be written as a product of disjoint cycles by induction, say (y 1... y l )(z 1... z m ). Since none of x i can appear in these cycles, we have g = hg = (x 1... x k )(y 1... y l )(z 1... z m ) is a product of disjoint cycles. Note that we think of S n as acting on the left. Therefore we multiply cycles as we compose maps, that is reading from right to left. For instance (1 2 3)(3 4 5) = ( ) as 3 is sent by (3 4 5) to 4 and then by (1 2 3) to 4, etc. A 2-cycle is also called a transposition. Lemma 3.2. Any element of S n can be written as a product of transpositions. Proof. By the previous lemma it is enough to show it for cycles. This is done by (x 1 x 2... x k ) = (x 1 x 2 )(x 2 x 3 ) (x k 1 x k ). Note however that this is in no way unique, not even the number of transposition is unique. Proposition 3.3. Two elements of S n are conjugate if and only if they have the same cycle structure. 15

16 Proof. : Let h = (x 1 x 2... x k )(y 1... y l ) be an element of S n written as a product of disjoint cycles. For any g S n, the permutation ghg 1 sends g(x 1 ) to g(x 2 ) as g(x 1 ) g 1 x 1 h x 2 g g(x 2 ). We see that ghg 1 = ( ) ( ) g(x 1 ) g(x 2 )... g(x k ) g(y 1 ) g(y 2 )... g(y l ) has the same cycle structure. In fact, we see that conjugation by g, just means applying g to the elements in the cycles. : Suppose h = (x 1 x 2... x k )(y 1... y l ) and h = (x 1 x 2... x k )(y 1... y l ) have the same cycle structure. Choose an element g S n that sends x 1 x 1, x 2 x 2,... x k x k, y 1 y 1,... Then ghg 1 = h by the computations in the first part of the proof. 3.1 The sign of permutations Let g S n and write k g for the number of orbits of the action of g on X. We say g is even and write sign(g) = +1 if n k g is even, otherwise g is odd and sign(g) = 1. For example when n = 7 and g = (1 2 3)(4 5), then k g = 4 and so g is odd. If g = 1 then it has n orbits and so 1 S n is even for all n. Any transposition is odd. More generally Lemma 3.4. A k-cycle is even if and only if k is odd. Proof. There is one orbit of size k and n k orbits of size 1. So it is even if and only if n (1 + n k) = k 1 is even. Lemma 3.5. Let g S n and let h S n be a transposition. Then sign(gh) = sign(g). Proof. Write h = (x y). Suppose first that x and y are in the same g -orbit. From the following picture (the action of g on the left and gh on the right hand side), we see that there is exactly one g -orbit that breaks into two gh -orbits; all others remain the same. Hence the sign changes. Next, we suppose that x and y are not in the same g -orbit. Again, we draw the action of g on the left and the one of gh on the right: 16

17 Therefore there are exactly two g -orbits that glue together to become a single gh -orbit. Again the sign changes. Proposition 3.6. If g S n is written as a product of k transposition, then g is even if and only if k is. Proof. As seen before 1 is even. Then each time we multiply by a transposition, the sign changes as shown in the previous lemma. Theorem 3.7. The map sign: S n {1, 1} is a homomorphism. Proof. Let g, h S n. We can write g as a product of k transpositions and h as a product of l transpositions. Then gh can be written as a product of k + l transpositions we if just multiply them. Hence sign(g) sign(h) = ( 1) k ( 1) l = ( 1) k+l = sign(gh). 3.2 The alternating group The kernel of the homomorphism sign: S n {±1} is called the alternating group A n. It is the set of all even permutations. It is a normal subgroup of index 2 in S n. Lemma 3.8. If n 3 then A n is generated by 3-cycles. Proof. Any element of A n can be written as an even number of transpositions. (w x)(y z) = (w z y)(w x y) and (x y)(x z) = (x z y) does the trick. Then Lemma 3.9. If n 5, then all 3-cycles are conjugate in A n. Proof. By proposition 3.3, for any two 3-cycles g and g there is an element th in S n such that hgh 1 = g. The lemma says there is a h in A n such that h gh 1 = g. So suppose h A n. Write g = (x y z). Since n 5, there are v and w not in {x, y, z}. Set k = (v w). Then kgk 1 = g gives (hk)g(hk) 1 = g with h = hk A n. Lemma 3.9 is wrong for n = 4. Check this! Corollary If n 5 and H A n contains a 3-cycle, then H = A n. Proof. If H contains a 3-cycle, then it contains all its conjugates because H is normal. By lemma 3.9, it contains therefore all 3-cycles and by lemma 3.8, we must have H = A n. Theorem If n 4, then A n is a simple group. 17

18 A group G is simple if it has no normal subgroup other than G and {1}. Proof. We have A 1 = A 2 = {1} and A 3 = C 3, so they are clearly cyclic. Suppose n 5 and let {1} H be a normal subgroup of A n. We want to show that H contains a 3-cycle as then corollary 3.10 implies H = A n. Pick 1 h H and let k be the length of the longest cycle in the decomposition of h into disjoint cycles. k 4: Say h = (w x y z... ). Set g = (w x y). Now h = ( ghg 1) h 1 belongs to H as H is normal. If v {w, x, y, z}, then h 1 (v) {w, x, y} and so h (v) = v. Also h (y) = y as y h 1 x g 1 w h x g h = (w x z). We conclude H = A n. y. So h H is a 3-cycle; in fact k = 3: If h has only one 3-cycle in its decomposition, then h = ( )( )( ) and hence h 2 is a 3-cycle and we have H = A n. Otherwise suppose h = (u v w)(x y z). Set g = (v w x) and consider ghg 1 h H. It is equal to (u y v x z) and hence by the case k 4, we have H = A n. k = 2: So h is a product of an even number of transpositions. Suppose first that h = (w x)(y z). Then we can pick n 5 we can pick a v {w, x, y, z}. Then ghg 1 h = (w x v) H and so H = A n. Finally suppose h = (s t)(u v)(w x)(y z). Then we take g = (t u)(v w) and ghg 1 h = (s w v)(t u x) H brings us back to the case k = 3 and we conclude H = A n. 18

19 4 Sylow s theorems Let G be a finite group. For every subgroup H, the order H divides G. Conversely: For what divisors of G is there a subgroup of this order? For instance there is no subgroup of order 6 in A 4. However for some divisors we can answer the question. Throughout this section p will be a fixed prime number. A finite group is called a p-group if its order is a power of p. Theorem 4.1. A non-trivial p-group has a non-trivial centre. Proof. The index of a centraliser C G (g) is either a power of p or 1; where the latter only happens when g Z(G). The class equation in (2.4) shows that p divides Z(G) as it divides G and all non-trivial indices of centralisers. Theorem 4.2. Let G be a p-group of order p m. Then there are subgroup 1 = H 0 < H 1 < < H m = G with H j = p j for all 0 j m. Proof by induction on m. If m = 1, the theorem is clear. Assume m > 1. By theorem 4.1, there is 1 g Z(G). Say g has order p k. Then h = g pk 1 has order p. Set H 1 = h, which is a normal subgroup in G as it is in the centre. Set Ḡ = G/H 1, which has order p m 1. By induction hypothesis, there are subgroups H 1 = 1 < H 2 < < H m = Ḡ with H j = p j 1. By the correspondence theorem, there are subgroups H j containing H 1 such that H j = H j /H 1. These groups satisfy the requirements of the theorem. We will refine this theorem a lot in theorem 6.13 in section 6.3. Let G be a finite group of order p m r with p not dividing r for some m 0. A subgroup P G is called a p-sylow (or p-sylow subgroup) of G if it is a p-group whose index in G is coprime to p. In other words, it is a subgroup of order p m. If G is a p-group then G is a p-sylow. If p does not divide G then {1} is a p-sylow. Otherwise it is not immediately clear that there exists a p-sylow in every group. Lemma 4.3. Let G be a finite group with a p-sylow P and let H < G. Then there is a g G and a p-sylow Q of H such that Q gp g 1. Proof. Let the group H act on the set X of left cosets for P in G by left multiplication. First, p can not divide #X = G : P. So there is at least one H-orbit whose size is coprime to p. Pick an element gp X in such an orbit and set Q = Stab H (gp ). By the orbit-stabiliser theorem 2.3, H : Q = # Orb H (gp ) is coprime to p. Then Q = { h H hgp = gp } = { h H g 1 hg P } = { h H h gp g 1} = H gp g 1 implies that Q is contained in the p-group gp g 1. Hence it is a p-group and so it is a p-sylow of H. Theorem 4.4 (First theorem of Sylow). Every finite group has a p-sylow. 19

20 Proof. By theorem 2.11, we can view the finite group G as a subgroup of GL n (F p ) for some n if F p denotes the field Z / pz with p elements. By the previous lemma 4.3, one only needs to show that GL n (F p ) has a p-sylow. Here it is: Set P to be the set of all upper triangular matrices with only 1s on the diagonal P = It is not difficult to show that P is a subgroup and to count that P = p n(n 1)/2 and GL n (F p ) = p n(n 1)/2 (p n 1)(p n 1 1) (p 1). Combining theorems 4.2 and 4.4, we obtain the following. Corollary 4.5. If p j divides G, then G has a subgroup of order p j. In particular, this proves a theorem of Cauchy Corollary 4.6 (Cauchy s theorem 1844). If p divides G, then G contains an element of order p. Now to a refinement of theorem 4.4 Theorem 4.7 (Sylow s theorems). Let G be a finite group. (a). Every subgroup in G which is a p-group is contained in a p-sylow of G. (b). All p-sylows of G are conjugate. (c). Let s p be the number of p-sylows of G. Then s p 1 (mod p) and s p divides the index of the p-sylows. Proof. Lemma 4.3 proves (a) directly. To prove (b), we apply lemma 4.3 with H and P being two distinct p-sylows of G. Then H gp g 1 and because they have the same size, we have H = gp g 1. Finally, we prove (c). Let G act on the set X of all its p-sylows by conjugation. The action is transitive by (b). Let P be a p-sylow. Its stabiliser is Stab G (P ) = N G (P ) =: H. The fact that P N G (P ) implies that s p = #X = G : H = G : P / H : P divides G : P. Note that P is normal in H = N G (P ) and hence it is the unique p-sylow of H by (b). We now consider the action of H by conjugation on X. The action is no longer transitive, for instance Orb H (P ) = {P } contains only one element. Let Q be another p-sylow of G and consider Stab H (Q). Since Stab H (Q) Stab G (Q) = N G (Q) and the latter has a unique p-sylow, Stab H (Q) can not contain P. Therefore # Orb H (Q) = H : Stab H (Q) is divisible by p for all Q P. We conclude that the action of H partitions X into one orbit {P } of size 1 and all other orbits of size divisible by p. Hence s p = #X 1 (mod p). { ( For example 1 ) } { ( 0 1 and 1 0 ) } 1 are two p-sylows of GL 2 (F p ). They are indeed conjugate by ( ) How many p-sylows are there in GL2 (F p )? 20

21 4.1 Applications Theorem 4.8. Let p > q be two distinct prime number and let G be a group of order pq. Then G is not simple. In fact G = P Q for a p-sylow P and a q-sylow Q. If furthermore p 1 (mod q), then G is cyclic. Proof. Let s p be the number of p-sylows of G. We have that s p must divide q, which means that s p is either 1 or q. Since we also have s p 1 (mod p), we conclude that s p = 1 because q < p can not be congruent to 1 modulo p. Let P be this unique p-sylow of G. It is normal and hence G is not simple. Let Q be any q-sylow. Then P is a cyclic group of order p and Q is a cyclic group of order q. If g belongs to P Q then its order must divide P = p and Q = q. This shows that P Q = {1}. Now the subgroup P Q has order P Q = pq = G. Therefore P Q = G which concludes the proof that G = P Q by Lemma 1.3. Assume now that p 1 (mod q). Let s q be the number of q-sylows. Then s q divides p and it is congruent to 1 modulo q. Hence s q = 1. So there is a unique q-sylow Q and it is normal. We have seen in an exercise in the first chapter that this shows that G = P Q. Hence G = C p C q = Cpq is cyclic. Note that the condition p 1 (mod q) is really necessary as we have constructed a non-abelian meta-cyclic group of order pq when p 1 (mod q) in section 2.5. Theorem 4.9. Let p and q be two distinct primes and suppose G is a group of order p 2 q. Then G is not simple. Proof. Let s p and s q denote the number of p-sylows and q-sylows in G. Assume that G is simple, so s p 1 and s q 1. We conclude as before that s p = q 1 (mod p), which also implies that p < q. This in turn exclude the possibility that s q = p. Now s q must divide p 2, which leave only s q = p 2 as a possibility. Now we reach a contradiction if we estimate the number of elements contained in p-sylows and q-sylows: Each q-sylow is cyclic of order q and two distinct such intersect in 1 only. So there are s q (q 1) = p 2 q p 2 = G p 2 non-trivial elements in them, none of which could appear as an element in a p-sylow. There are at least two p-sylows of order p 2, so the number of elements in them must be larger than p 2, which is impossible. With similar sort of techniques, it is possible to show that no group whose order is a product of three distinct primes is simple. With much more work of the same kind one can give the list of all simple groups of order less than 100: Namely, there are only the cyclic groups C p for primes p and the alternating group A 5 of order

22 5 Finitely generated abelian groups In this section all groups are abelian and we write the operation as + and the neutral element as 0. Recall that an abelian group A is finitely generated if there is a set {a 1, a 2,..., a n } such that every element a A can be written as a Z-linear combination a = x 1 a 1 +x 2 a x n a n for some x 1,..., x n Z. If the integers x i are unique, then A is said to be free on {a 1,..., a n }. The change of the generating set is a bit like a change of basis in linear algebra. Suppose b 1 = p 11 a 1 + p 12 a p 1n a n b 2 = p 21 a 1 + p 22 a p 2n a n.. b n = p n1 a 1 + p n2 a p nn a n for some p ij Z. Lemma 5.1. If the matrix P = (p ij ) has determinant ±1, then {b 1, b 2,..., b n } is also a generating set for A. Moreover if A was free on {a 1, a 2,..., a n }, then so it is on {b 1, b 2,..., b n }. Proof. If det(p ) = ±1, then P 1 has also integer coefficients. Hence each a i can be expressed uniquely as a Z-linear combination of the b j. Therefore any a A can be written as a combination of the b j ; and this expression is unique if A were free on {a 1,..., a n }. Recall that the group of all integer matrices with determinant ±1 is denoted by GL n (Z). 5.1 Free abelian groups Another way of saying that A is free on {a 1,..., a n } is to say that the map A Z n sending a = x 1 a x n a n to (x 1,..., x n ) is an isomorphism. Lemma 5.2. If A is free on {a 1, a 2,..., a r } and free on {b 1, b 2,..., b s }, then r = s. The number r is then called the rank of A. Proof. A / 2A is isomorphic to (Z / 2Z ) r and isomorphic to ( Z / 2Z ) s. Comparing the size gives r = s. Theorem 5.3. Any subgroup B of a finitely generated abelian group A of rank r is free of rank at most r. Proof by induction on r. If r = 1, then the only subgroups of A = Z are {0}, which is free of rank 0, and mz for some m > 0 and they are free of rank 1. Say A is free on {a 1,..., a r } with r > 1. Consider the map ϕ: A Z that sends x = x 1 a x r a r to x 1. Then ker ϕ is a free abelian group of rank r 1. By induction, its subgroup B = B ker ϕ is free of rank at most r 1. Now, if ϕ(b) = {0}, 22

23 then B = B and we are done. So suppose ϕ(b) = mz for some m > 0. Pick a c B such that ϕ(c) = m. For every b B, there is a k Z such that ϕ(b) = km. Therefore the element b = b k c belongs to B = B ker ϕ. Hence any element b B can be written in a unique way as a sum b = b +k c with k Z and b B. So if we add c to the generating set of B, we find a generating set for B on which it is free of rank at most r. 5.2 Smith normal form Theorem 5.4. Let M be an m n matrix with integer coefficients. Then there are matrices P GL m (Z) and Q GL n (Z) and a sequence of integers d 1, d 2,..., d t such that d i divides d i+1 for all 1 i < t and such that d 1 d 2 0 d 3 0. P M Q = 0.. dt 0 0 is a matrix with its only non-zero coefficients on the diagonal. Proof. Among the elements in GL n (Z) and in GL m (Z) there are those giving us permission to do certain row and column operations: We may multiply a row (or a column) by 1. We may swap two rows (or two columns). We may add an integer multiple of one row (or column) to another. For instance the last is done by multiplying with the matrix P having 1s on the diagonal and a single non-zero integer values off the diagonal. It is important to remember that we are never allowed to divide. Our aim is to bring M into the requested form by applying these three operations on row and columns. First we may swap rows and columns (and the sign if needed) to have a positive (preferably small) value a in the top left hand corner. If this were not possible then the matrix is the zero-matrix and we are done. If there is a row, say the i-th row, such that its first entry is not divisible by a, then we can add a multiple of the first row to the i-th row to make sure that the first entry of the i-th row is now an integer between 0 and a. Then we swap it with the first row. In this manner using rows, but in a similar fashion by using columns, we make the top left entry a > 0 smaller. We will only be able to do this a finite number of times and then we will have reached the situation where all entries in the first row and in the first column are divisible by a. 23

24 Now we can obtain 0 in all the entries of the first row except the top entry by adding multiples of the first row to the other rows and similar for the first column. We have now reached a matrix of the form a M 0 for some matrix M. We repeat the process with M and so forth. Eventually we reach a zero matrix. Therefore, we will arrive at a matrix of the form a b and all that remains to do is to make sure the divisibility holds. This can be done with lemma 5.5 below. First an example for the process so far with ( ) M = Our first step is to bring the 2 in the top left corner (the so-called pivot ). ( ) Next, we see that the second column has a top entry that is not divisible by 2, so we subtract the first column from it and then swap the two first columns. ( ) Now, we have reached the stage where all the first entries in the columns and rows are divisible by the top left entry a = 1. We may subtract 13 times the first row from the second, twice the first column from the second and 5 times the first column from the last column. I will also switch the signs of the second and last column at the end. ( 1 0 ) So we reduced the problem to the smaller matrix M = (32 76). The pivot is already there, we we start by subtracting twice the second column from the first and then swap the two to obtain (12 32). We do exactly the same again and have (8 12). Again we subtract the first column from the second and swap them (4 8). This time all the entries in the (top) 24

25 Group Theory G13GTH cw 12 row are divisible by b = 4. We can get a 0 by subtracting twice the first column from the second. Finally, we reached ( ) which is by chance already in the desired Smith normal form. Note that the operations done to get from (32 76) to (4 0) is just the Euclidean algorithm computing the greatest common divisor gcd(32, 76) = 4. Lemma 5.5. The matrix ( ) ( a 0 0 b can be transformed to g 0 ) 0 l with g = gcd(a, b) and l = lcm(a, b). Proof. First we may suppose b > a > 0. We start by adding the bottom row to the top row to obtain ( ) a b 0 b. Now we subtract the first column from the second as often as to reduce the top right entry to 0 b < a. Now we subtract the second from the first column to get the top left entry a to be smaller than b. Now the matrix looks like ( ) a b kb b for some k Z. And so forth, we see that we are just doing a Euclidean algorithm with the top row. It will reach (after a column swap if necessary) a matrix where the top row is (g 0) with g = gcd(a, b). Both entries in the bottom rows are multiples of b. Since g divides b, we can subtract a multiple of the first row from the second to achieve a zero in the bottom left entry. Now the matrix looks like ( ) g 0 0 l. Since the determinant never changed, we must have gl = ab and hence l = lcm(a, b). We can also illustrate the lemma with an example. ( ) ( ) r 1 r 2 +r ( ) c 2 c 2 c c 1 c 2 ( ) ( ) c 2 c 2 2c ( ) r 2 r 2 3r ( ) c 2 c Theorem 5.6. Let ϕ: F F be a homomorphism between two finitely generated free abelian groups. Then there are generating sets {e 1,..., e n } of F and {f 1,..., f m } of F and integers d 1, d 2,..., d t such that ϕ(e i ) = d i f i if 1 i t and ϕ(e i ) = 0 if i > t and such that d i divides d i+1 for all 1 i < t. This is just a reformulation of the Smith normal form. Take any generating sets for F and F on which they are free. Then write the the matrix for ϕ with respect to these generating set (as we do for linear maps in linear algebra) by writing the coefficients of the images of the generating set of F as columns. The operations described in the proof of the Smith normal form are just changes of generating sets. 25

26 5.3 The fundamental theorem on finitely generated abelian groups Theorem 5.7. Let A be a finitely generated abelian group. Then there exists a unique r 0 and integers 1 < m 1 m 2 m t such that m i divides m i+1 for all 1 i < t and such that A = Z / m1 Z Z / m2 Z Z / mtz Z r. Proof. Choose a generating set {a 1,..., a m } for A. Let F be a free abelian group of rank n generated by {f 1,..., f m } and consider the map ψ : F A sending f i to a i for all 1 i m. Let R be the kernel of ψ. By theorem 5.3, R is a free abelian group of rank n m. Now apply theorem 5.6 to the inclusion map ϕ: R F. So there is a generating set {f 1,... f m} of F such that R, the image of ϕ, is equal to d 1 Z f 1 d 2Z f 2 d sz f s for some integers d i with the usual divisibility and s n. Since A = F/R, we find A = Z / d1 Z Z / d2 Z Z / dsz Z m s. So we set r = m s and delete the d i = 1 from the sequence. In practice the group A could be given by generators and relations. For instance, suppose A is generated by three elements {a 1, a 2, a 3 } subject to the relations 3 a 1 2 a a 3 = 0 and 7 a a 2 11 a 3 = 0. Then the matrix for ϕ, called the relation matrix, is 3 7 M = We have computed the Smith normal form for the transpose of this matrix before and found d 1 = 1 and d 2 = 4. We conclude that A = Z / 4Z Z. The example transforming ( ) ( to 1 0 ) 0 6 can now be viewed as another proof that Z / 2Z Z / 3Z = Z / 6Z. Corollary 5.8. Let A be a finitely generated abelian group. Then there are (not necessary distinct) prime numbers p 1, p 2,..., p s and integers n 1, n 2,..., n s 1 such that A = Z / n p 1 1 Z Z / n p 2 2 Z Z / p ns s Z Z r An element of finite order in an abelian group is also called torsion. It is easy to see that the set of all torsion elements in A is a subgroup denoted by Tors(A). For a finitely generated abelian group A as in theorem 5.7, this is simply the finite group Tors(A) = Z / m1 Z Z / m2 Z Z / mtz. Corollary 5.9. If a finitely generated abelian group has no torsion elements other than 0, then it is free. This, by the way, is the starting point of another, quite different proof of the main theorem 5.7 which is given in G13RIM for general principal ideal rings. 26

27 6 Series Let G be a group. A series for G is a finite sequence of subgroups 1 < H 1 < H 2 < < H n = G. The left hand 1 is a short hand notation for the trivial subgroup {1} < G. The natural number n is called the length of the series. If H i H i+1, then H i+1 /H i is called a factor of the series. 6.1 Composition series A series is called a composition series if H i H i+1 for all 1 i < n and the factors H i /H i+1 are simple group. In other words H i+1 H i and there is no larger normal subgroup H i < N H i+1. We write 1 H 1 H 2 H n = G but recall that this does not mean that H 1 G if n > 2. Lemma 6.1. Every finite group has a composition series. Proof. If G is simple then 1 < G is a composition series. Otherwise take a normal subgroup H G of maximal order. Then take a normal subgroup of maximal order in H etc. Because G is finite, this will eventually stop. Here a few examples of composition series that also show that there is not a unique such. 1 C 2 = (1 2)(3 4) C 2 C 2 = (1 2)(3 4), (1 3)(2 4) A 4 S 4 1 C n D n 1 C p C p C p C p C p C p if p is prime 1 C 2 C 4 C 12 1 C 2 C 6 C 12 1 C 3 C 6 C 12 Proposition 6.2. If G has a composition series of length n and N G, then N has a composition series of length at most n. Proof. Let 1 H 1 H n = G be a composition series of G. Now consider 1 (H 1 N) (H 2 N) (H n N) = N which is a sequence of subgroups of N. We will show that at each step we have either equality or a simple quotient; hence by deleting repetition, we have a composition series of length at most n. 27

28 By the third isomorphism theorem, we have H i+1 N / H i N = H i+1 N / H i (H i+1 N) = H i (H i+1 N) / H i. This is a normal subgroup of H i+1 /H i as both H i and H i+1 N are normal in H i+1. Hence it is either trivial or all of H i+1 /H i because this factor is simple. If it is trivial, then H i+1 N = H i N. Otherwise (H i+1 N)/(H i N) is simple. Theorem 6.3 (Jordan-Hölder). Any two composition series for a group G have the same length and equal set of factors (up to permutation). Proof by induction on the length n of the shortest composition series. If this length is n = 1, then G is simple and there is only one composition series. Assume the G has two decomposition series 1 H 1 H 2 H n 1 H n = G (6.1) 1 J 1 J 2 J m 1 J m = G (6.2) of lengths m n > 1 respectively. Set H = H n 1 and J = J m 1. First, if H = J, then the two series are composition series of length n 1 of H and so by induction the theorem holds. G J H J = J m 1 H = H n 1 K = H J Therefore we may assume that H J. Since H < H J is normal in G but G/H is simple, we have H J = G. Set K = H J. By the third isomorphism theorem, we have H / K = HJ / H = G/ H and J / K = G/ J and they are both simple. By proposition 6.2, the subgroup K < H has a composition series of length at most n 1, say J m 2 H n 2 1 K 1 K 2 K r = K K r 1 with r < n. By the above argument, 1 K 1 K 2 K r 1 K H G (6.3) 1 K 1 K 2 K r 1 K J G (6.4) are both composition series of G. In particular, we have now two composition series (6.1) and (6.3) of H. By induction they have the same length, i.e. n 1 = r + 1. Similarly for J and we obtain m 1 = r + 1. Therefore n = m. Finally the factors of (6.3) and (6.4) are the same apart from the swapping of the last two. Comparing (6.1) and (6.3) for H, we know by induction that the factors for H are K 1, K 2 /K 1,..., K/K r 1, H/K = G/H. Therefore the set of factors for G in (6.1) and (6.2) is {K 1, K 2 /K 1,..., K/K r 1, G/H, G/J} for both composition series. This theorem shows that the simple finite groups are the building blocks of finite groups. However a finite group G is not determined by the set of simple composition factors. For instance 1 C 2 C 6 and 1 C 2 S 3 have both the factors {C 2, C 3 }. The question of classifying all finite group amounts to classify all finite simple groups and to understand completely all ways of how they can be put together (extension problem). 28

29 6.2 Soluble groups A group G is soluble if it has a series 1 H 1 H 2 H n = G such that H i+1 /H i is abelian for all 0 i < n. Proposition 6.4. Every subgroup and every quotient of a soluble group is soluble. Conversely, if N G is such that N is soluble and G/N is soluble, then G is soluble. Proof. Let G be a soluble group and let 1 H 1 H n = G be a series with abelian factors. If J < G, consider whose factors are abelian because 1 H 1 J H 2 J H n J = J H i+1 J / H i J = H i (H i+1 J) / H i H i+1 / H i. Therefore by deleting repetition, we show that J is also soluble. Now assume N G. Set H i = (H i N)/N. Then 1 H 1 H 2 H n = G/N has also abelian factors because the second and third isomorphism theorems imply H / i+1 Hi = H i+1 N / H i N = H i+1 (H i N) / H i N = H / i+1 H i+1 (H i N) ( ) = Hi+1 /H i ) /(Hi+1 (H i N))/H. i Therefore G/N is soluble. Finally let G be a group and suppose N G is soluble with G/N soluble, too. We have series with abelian factors for both of them 1 N 1 N 2 N n = N 1 N 1 N 2 N m = G/N By the correspondence theorem, there is a subgroup N n+i < G such that N N n+1 and N n+i /N = N i. It is easy to check that 1 N 1 N 2 N n N n+1 N n+m = G is a series with abelian factors. Therefore G is soluble. Corollary 6.5. Every finite p-group is soluble. Proof by induction on its order. It is clear if G has order p. Otherwise the centre Z(G) is non-trivial by theorem 4.1. Then Z(G) is soluble and G/Z(G) is soluble by induction. The previous proposition shows that G is soluble. 29

30 The terminology comes from Galois theory. Galois was also the first to use the term group for a permutation group. His main theorem says that a polynomial has a soluble Galois group if and only if it is solvable by radical. (See G13GNF) Let G be a group. We define the derived subgroup G = [G, G] as the subgroup of G generated by the commutators [g, h] = ghg 1 h 1 for g, h G. Because g[h, h ]g 1 = [ ghg 1, gh g 1], the derived group is normal in G. REcall the following lemma whose proof is an exercise. Lemma 6.6. Let N G. Then G/N is abelian if and only if G N. Inductively we define a sequence of subgroups G (3) G (2) G (1) = G G (6.5) by setting G (i+1) = [ G (i), G (i)]. In particular G (i) = 0 if and only if G (i 1) is abelian. For example the derived subgroup of S n is A n and the derived subgroup of A n is itself if n > 4. So this example shows that the sequence (6.5) need not to terminate. Theorem 6.7. A group G is soluble if and only if there is an r 1 such that G (r) = {1}. Proof. : Using lemma 6.6, we can show that 1 = G (r) G (r 1) G G becomes a series with abelian quotients by deleting repetitions. : Let 1 H 1 H n = G be a series with abelian factors. We will show that G (i) H n i by induction on i. This will show that G (n) = {1}. First, if i = 0, then G (0) = G = H n. Next, G (i+1) = ( G (i)) ( Hn i ) by induction. Since H n i /H n i 1 is abelian, ( H n i ) H n i 1 again using lemma 6.6. Therefore G (i+1) H n (i+1). Moreover, the proof shows that for a soluble group G, is the shortest series with abelian quotients. 6.3 Nilpotent groups 1 G (r 1) G (2) G G A group G is called nilpotent if it admits a series 1 N 1 N 2 N n = G (6.6) such that N i G and N i+1 /N i Z ( G/N i ). Such a series is called a central series. For any subgroups H and J of a group G, we define [H, J] to be the subgroup generated by [h, j] with h H and j J. Lemma 6.8. Let N G and N H G. [H, G] N. Then H/N Z ( G/N ) if and only if 30

31 Proof. To say that H/N Z ( G/N ) is equivalent to ask that hgn = ghn for all h H and g G. This is also the same as to ask that hgh 1 g 1 N for all h H and g G, which is exactly the right hand side. Inductively we define the higher commutator subgroups D i (G) by D 0 (G) = G and D i+1 (G) = [D i (G), G]. The lower central sequence of G is 1 D 3 (G) D 2 (G) D 1 (G) D 0 (G) = G. (6.7) Of course we have D 1 (G) = G. But in general G (i) D i (G) may be a strict inclusion. Proposition 6.9. A group G is nilpotent if and only if there is an r 1 such that D r (G) = {1}. Proof. : Note first that D i (G) G for all i. Then by lemma 6.8, and the fact that D i (G) = [D i 1 (G), G] gives that D i 1 (G)/D i (G) lies in Z ( G/D i (G) ). Hence the lower central sequence is a series as in (6.6). : Let 1 N 1 N n = G be a series as in (6.6). We will prove by induction that D i (G) N n i. First for i = 0, we have D 0 (G) = G = N n. Next, we have D i+1 (G) = [D i (G), G] [N n i, G] N n i 1 by the induction hypothesis and lemma 6.8. The higher centre Z i (G) is defined by Z 0 (G) = {1} and Z i+1 (G) is the unique subgroup in G such that Z i+1 (G) / ( Z i (G) = Z G / ) Z i (G). (6.8) The existence and uniqueness of this subgroup is guaranteed by the correspondence theorem. By a slight variant of lemma 6.8, the condition (6.8) is the same as to ask that [ Z i+1 (G), G ] = Z i (G). For instance Z 1 (G) = Z(G). This leads to an increasing sequence of groups called the upper central sequence. 1 Z 1 (G) Z 2 (G) Theorem A group G is nilpotent if and only if there is an r 1 such that Z r (G) = G. If so, then the length of the upper and lower central series are equal. The common length of the upper and lower central series are called the nilpotency class. Proof. : If Z r (G) = G, then the upper central series is a central series as in (6.6). : Let 1 N 1 N n = G be a central series. We prove by induction that N i Z i (G). First for i = 1, we have N 0 = {1} = Z 0 (G). Next, by the definition of a central series together with lemma 6.8, we have [ N i+1, G ] N i, which is contained in Z i (G) by induction hypothesis. So again by lemma 6.8, N i+1 /Z i (G) is contained in the centre of G/Z i (G). This means that N i+1 Z i+1 (G). Suppose now that G is nilpotent. Let r be the length of the lower central series and s be the length of the upper central series. Then the above proof of shows that D r i (G) Z i (G). Hence Z r (G) = G implies s r. The proof of in proposition 6.9 shows that D i (G) Z s i (G) and hence D s (G) = {1}, which implies r s. Therefore r = s. 1 I changed the notation by shifting the index by one. Books will always take D 1(G) = G. 31

32 Figure 2: Central series are between the upper and lower central series The proofs of in this theorem 6.10 and in proposition 6.9 shows that for every nilpotent group of class r and any central series 1 N 1 N n, we have D n i (G) N i Z i (G). In other words Z i (G) is the fastest increasing and D r i (G) the fastest decreasing central series of G. See figure 2 for a picture of how one could think of this. Lemma If ϕ: G H is a surjective homomorphism, then ϕ ( D i (G) ) = D i (H). The proof is an exercise. Proposition Let G be a nilpotent group of class r and H G and N G. Then H and G/N are both nilpotent of class at most r. Proof. Because D i (H) D i (G), it is clear that D r (H) = {1}. The D i (G/N) is the image of D i (G) under G G/N by lemma Hence D r (G/N) = {1}. It is important to remark that the converse is not true (unlike for solubility in proposition 6.4). For instance C 2 and C 3 are both nilpotent (of class 1), but S 3 is not nilpotent. The following is a refinement of corollary 6.5 and theorem 4.2. Theorem Every finite p-group is nilpotent. Proof by induction on the order. If G is of order p then it is cyclic and hence it is nilpotent of class 1. Otherwise, we know that the centre Z(G) is non-trivial by theorem 4.1. By induction G/Z(G) has a central series, say 1 N 2 /Z(G) N 2 /Z(G) N n /Z(G) = G/Z(G). Then 1 N 1 = Z(G) N 2 N n = G 32

33 is a central series for G as N i+1 /N i = ( Ni+1 /Z(G) ) / (Ni /Z(G) ) Z(( G/Z(G) ) /(Ni /Z(G) )) = Z ( G /N i ). Lemma If G and H are nilpotent groups then G H is nilpotent, too. Yet again, the proof is an exercise. Theorem Let G be a finite group. Then the following are equivalent. (a). G is nilpotent. (b). If H < G then H N G (H). (c). All maximal subgroups of G are normal. (d). All Sylows of G are normal. (e). ab = ba for all a, b G whose orders are coprime. (f). G is the direct product of its Sylows. Proof. (a) (b) : Let H < G. Let n be the largest such that Z n (G) H, which exists by theorem 6.10 because G is nilpotent. So there is a g Z n+1 (G) which does not belong to H. Now for all h H, we have that ghg 1 = [g, h]h belongs to Z n (G) H = H as [ Z n+1 (G), G ] = Z n (G). Hence g N G (H) but g H. (b) (c) : If H is a maximal subgroup, then H < N G (H) G by (b). Hence N G (H) = G and H is normal. (c) (d) : Let P be a p-sylow for some prime p dividing the order of G. If N G (P ) = G, then P is normal. We will assume now that N G (P ) G. Then there is a maximal subgroup H containing N G (P ). By (c), H G. Take a g G which does not belong to H. Because H is normal gp g 1 is still contained in H. Therefore P and gp g 1 are both p-sylows of H. By theorem 4.7(b), there is a h H such that h(gp g 1 )h 1 = P. In other words hg N G (P ) H. This implies that g H which is a contradiction. (d) (f) : We prove this by induction on the number of primes dividing the order of G. First, if G is a p-group, then (f) is true. Now let p be a prime dividing the order of G and let P be the unique p-sylow of G. Let H be the subgroup in G generated by all q-sylows with q p. By induction H is the direct product of all of them and we find that H is the product of the orders of them, which gives H = G / P. No element other than 1 can have order dividing p in H, so H P = {1}. Therefore HP = H P = G shows that HP = G. For all h H and g P, the commutator [h, g] is in P as it is (hgh 1 )g 1 and P is normal. It also belongs to H because H is normal and [h, g] = h(gh 1 g 1 ). Therefore [h, g] P H = {1} shows that elements in H and P commute. We conclude by theorem 1.1 that G = H P. So G is the direct product of all its Sylows. (f) (a) : This is simply the combination of theorem 6.13 and lemma (f) (e) : Write G = P 1 P 2 P n where P i are the non-trivial Sylow groups of G. Write a as (a 1, a 2,..., a n ) with a i P i and similarly for b = (b 1, b 2,..., b n ) under this isomorphism. If p is a prime dividing G and P i the p-sylow of G, then at most one of a 33

34 and b can have order divisible by p. This means that a i = 1 or b i = 1. Hence a i b i = b i a i and so a and b commute. (e) (d) : Let p be a prime dividing G and let P be a p-sylow of G. For any element g G of order coprime to p, the conjugate gp g 1 = P by (e). Hence N G (P ) contains all q-sylow of G with q p and it contains P, too. Since the Sylows for different primes intersect in {1}, we obtain that N G (P ) G showing that N G (P ) = G and hence that P is normal. In fact the equivalence of (a), (b) and (c) does not need the asumption that G is finite. 34

35 7 Primitive actions Let G be a group acting on a set X. For any k 1, we say that the action is k-transitive if for any k distinct elements x 1, x 2,..., x k and any k distinct 2 elements y 1, y 2,..., y k there exists a g G such that g x i = y i for all 1 i k. If furthermore this g G is unique in all cases, then we say that the action is sharply k-transitive. In particular a 1-transitive action is just a transitive action and a sharply 1-transitive action is a regular action. Lemma 7.1. Let G act transitively on X and let x X. Then G acts [sharply] k- transitively on X if and only if Stab G (x) acts [sharply] (k 1)-transitively on X \ {x}. Proof. : Take k 1 distinct elements x 1, x 2,..., x k 1 in X \ {x} and another k 1 distinct elements y 1, y 2,..., y k 1. Then there is a [unique] g G sending x i to y i and x to x because the action is [sharply] k-transitive. Therefore g Stab G (x) shows that Stab G (x) acts [sharply] k 1-transitively on X \ {x}. : Let x 1, x 2,..., x k and y 1, y 2,..., x k be two sets of distinct elements in X. Since the action is transitive, we find a g G such that g x k = x and a g G such that g y k = x. By hypothesis there is a [unique] element h Stab G (x) such that h (g x i ) = g y i for all 1 i k 1. Now g 1 hg sends x i to y i for all i including k. Corollary 7.2. The action of the symmetric group S n on {1, 2, 3,..., n} is sharply n- transitive. For n 3, the action of A n is sharply (n 2)-transitive. A deep theorem of Jordan shows that if k > 5, then S k and A k+2 are the only two examples of a sharply k-transitive action. For k = 4 and 5, there is one more sharply k- transitive action given by the so-called Mathieux groups: The simple group M 12 of order has a sharply 5-transitive action on a set of 12 elements and the simple group M 11 of order 7920, the smallest of the sporadic simple groups, has a sharply 4-transitive action on a set of 11 elements. Using the classification of finite simple groups it is possible to classify all actions that are k-transitive with k 2. Let G act on a set X. A block Y is a subset of X with Y X and #Y > 1 such that, for all g G, the set g Y := { g y y Y } is either equal to Y or disjoint to it. If the action is not transitive, then an orbit with more than one element is a block. Here is an example of blocks which are not orbits: The group G of isometries of the cube acts transitively on the eight vertices of the cube. Let Y be the set containing one of these points together with the one diagonally opposed to it. Then Y is a block. In general, let G act on a set X and let Y be a block. Then all g Y are blocks, too, and X is partitioned into equal sized blocks. This will be proven in an exercise. The action of a group on a set X is called primitive if it is transitive and there are no blocks in X. Proposition 7.3. If the action of G on X is 2-transitive, then it is primitive. 2 x 1 = y 3 is allowed by y 1 = y 3 not 35

36 Proof. Let Y X with #Y > 1. Pick x Y and y Y. There is a g G such that g x = y and g y Y. Hence Y is not a block. However the converse is not true. Let X be the set of subsets of {1, 2,, n} with 2 elements. Consider the natural action of S n on X. It is an exercise to show that this action is primitive but not 2-transitive. Theorem 7.4. Let G act transitively on X. The action is primitive if and only if Stab G (x) is a maximal subgroup of G for any x X. Proof. Let x X. We actually prove that there is a subgroup Stab G (x) < H < G if and only if the action is not primitive. : Given such a group H, we consider Y := { h x h H }. Since Stab G (x) < H, there are at least two elements in Y. Let g G such that g Y Y. Then there are y = h x and y = h x in Y with h, h H such that g y = y. Therefore h 1 gh x = h 1 y = x implies that h 1 gh Stab G (x) < H and hence that g H. So if g Y Y, then g Y = Y. Since H < G, this shows that Y X. Therefore Y is a block and the action is therefore not primitive. : Suppose there is a block x Y X. Set H = { g G g Y = Y }. It is not difficult 3 to show that H is a subgroup of G. As the action of G on X is transitive, there exist an element in G that does not belong to H. Let g Stab G (x). Since g x = x Y g Y, we must have g Y = Y. Hence Stab G (x) H. There are at least two elements x and y in Y. Let g G such that g x = y, which exists because the action is transitive. Again g Y Y and so g H. However g Stab G (x). Therefore Stab G (x) < H < G shows that the stabiliser is not a maximal subgroup. Theorem 7.5 (Iwasawa). Let G act faithfully and primitively on a set X and let x X. Suppose there is an abelian subgroup A Stab G (x) such that {gag 1 } g G generates all of G. Then any normal subgroup {1} N G contains the derived subgroup G = [G, G]. Proof. Set H = Stab G (x). By the previous theorem H is a maximal subgroup. Let {1} N G. If N were contained in H, then it would be contained in g G ghg 1 = ker ρ = {1} because the action is faithful. By the maximality of H the subgroup NH must be equal to G, meaning that every g G can be written as nh for some n N and h H. Then gag 1 = nhah 1 n 1 = nan 1 since A H. Because N G, we have nan 1 NA. As the gag 1 generate all of G, we must have NA = G. Finally G / N = NA/ N = A/ A N shows that G/N is abelian, which means that G N by lemma 6.6. Theorem 7.6. The alternating group A 5 is simple. 3 One can also view it as the stabiliser of Y of the action of G on the set of blocks partitioning X 36

37 Proof. The group A 5 acts faithfully and 2-transitively on {1, 2, 3, 4, 5}. The stabiliser of 5 is isomorphic to A 4, which contains an abelian normal subgroup V 4. Now g G gv 4g 1 is the conjugacy class of elements of the form (..)(..) in A 4. Because (1 2)(3 4)(1 2)(4 5) = (3 4 5) and so forth all 3-cycles, and hence by lemma 3.8 all of A 5, are generated by elements of this form. Therefore the theorem applies and any normal subgroup 1 N A 5 contains [A 5, A 5 ] = A 5. Therefore A 5 is simple. Theorem 7.7. The alternating group A n is simple for all n > 5. Proof. We prove this by induction. The anchor step is proved is the previous theorem. Let G = A n act on X = {1, 2,..., n}. For any x X, the stabiliser Stab G (x) is isomorphic to A n 1, which is simple by induction hypothesis. Suppose now 1 N A n. Step 1: The action of N on X is free. Let x X. Consider Stab N (x) = Stab G (x) N, which is a normal subgroup of Stab G (x). If Stab N (x) = Stab G (x), then Stab G (x) N. However the action of G is primitive and so Stab G (x) is maximal; hence Stab G (x) = N, meaning that A n 1 A n, which is not the case. Since Stab G (x) is simple, this leaves only the possibility that Stab N (x) = {1} and hence that the action of N is free. Step 2: The action of N on X is transitive. Let Y be an orbit of the action by N on X. By step 1, #Y > 1. Suppose g G sends an element x Y to an element Y. Then there is a n N such that g x = n x. Now given any other element y Y, say with y = n x for n N, we find that g y = gn x = gn g 1 g x = gn g 1 n x lies again in Y, since gn g 1 and n both belong to N G. We have therefore shown that g Y is either Y or disjoint from it. however the action of G is primitive on X, so Y can not be a block. Hence Y = X shows that N acts transitively and hence by step 1 regularly on X. Step 3: G = N H for H = Stab G (x) for any x X. We have shown in step 1 that H N = {1}. Let g G. Then, by step 1 and 2, there exists a unique n N such that g x = n x. Therefore n 1 g = h Stab G (x) = H. We have shown that every element in G can be written uniquely as nh. Furthermore, the action of H on N by conjugation identifies to the action of H on X via the bijection N X given by n n x, because hnh 1 x = h n x for all h H and n N. Step 4: N is abelian. Since the action of G is 2-transitive, the action of H on X \ {x}, and hence on N \ {1} by conjugation, is transitive. Since all non-trivial elements in N are conjugate in G, they must all have the same order. Furthermore this order must be a prime number p, otherwise some power of any element would have a different order. Hence N is a p-group and has therefore a non-trivial centre Z(N). Let z Z(N), h H and n N. Then ( hzh 1 ) n = hz ( h 1 nh ) h 1 = h ( h 1 nh ) zh 1 = n ( hzh 1) proves that hzh 1 Z(N). Since the action of H is transitive on non-trivial elements N must be abelian. Step 5: Final contradiction. As n 6, we have that H = A n 1 is at least 3-transitive on X \ {x} and on N \ {1}. Pick two distinct n, m N \ {1} with n m 1. Note that 37

38 this is possible because #N = #X > 3. Then we find a h H such that hnh 1 = n and hmh 1 = m, but hnmh 1 hnmh 1, which is absurd. The O Nan-Scott theorem classifies all primitive actions of S n, i.e. it classifies all maximal subgroups of S n. 38

39 8 Projective linear group Let K be a field and n 1. On the set K n+1 {(0, 0, 0,..., 0)} we define an equivalence relation by (x 0, x 1,..., x n ) (y 0, y 1,..., y n ) if and only if there is a λ K such that x i = λy i for all i = 0, 1,..., n. The set of equivalence classes is called the projective n-space and denoted by P n (K). The class containing (x 0, x 1,..., x n ) is denoted by (x 0 : x 1 : : x n ). Hence we have for all λ K that (x 0 : x 1 : : x n ) = (λx 0 : λx 1 : : λx n ). Here is how we visualise the projective space. P n (K) is the set of all lines in K n+1 going through the origin. If we take any hyperplane in K n+1 that does not go through the origin, then most of the lines through the origin will meet this hyperplane. Concretely, we can take the hyperplane x 0 = 1. We identify now K n, the traditional euclidean n-space, with a subset of P n (K) by K n P n (K) (x 1, x 2,..., x n ) (1 : x 1 : x 2 : : x n ). Note that the image is the subset { (x 0 : x 1 : : x n ) x 0 0 } as these are exactly the ones that can be scaled with a λ K to have x 0 = 1; in other words these are the line that are not parallel to the hyperplane x 0 = 1. The complement of K n in P n (K) can be identified with a copy of P n 1 (K). In particular, the projective line P 1 (K) is just an euclidean line K plus a point (0 : 1) at infinity. The projective plane P 2 (K) is an euclidean plane K 2 together with a projective line at infinity. 0 Figure 3: Euclidean and projective 2-space Any d-dimensional projective subspace in P n (K) is the set of lines belonging to a (d+1)- dimensional vector subspace of K n+1. For instance a projective line in P 2 (K) is the set of lines in K 3 belonging to the same vector space of dimension 2. Now each such plane through the origin will intersect our x 0 = 1 in a line, except for the one with equation x 0 = 0. In other words the projective lines of P 2 (K) are exactly the lines of K 2 together with the line at infinity. One of the advantages of projective geometry is that any two lines in P 2 (K) intersect in exactly one point: Indeed any two vector spaces of dimension 2 in K 3 intersect in a line 39

40 going through the origin. With respect to the subset K 2 P 2 (K), it means that any two parallel lines in K 2 intersect now exactly once on the line at infinity. So we can think of P 2 (K) as K 2 together with a point at the end of each direction of a set of parallel lines. In projective coordinates any line can be described by an equation a x 0 + b x 1 + c x 2 = 0 for some a, b, c not all equal to zero. In fact the set of all lines in P 2 (K) is the set of all (a : b : c), which forms itself a copy of P 2 (K). This is an instance of a very nice duality holding in general for projective geometry. Example over a finite field. Let K = F 2, the field with 2 elements. Then P 1 (F 2 ) has 3 elements and so P 2 (F 2 ) has = 7 elements. The best way to draw them is done in figure 4: Figure 4: The projective plane over F 2 The group GL n+1 (K) acts on K n+1 and it sends line through the origin to line through the origin. In other words it acts on P n (K). The action is no longer faithful, because the matrix ( λ 0 0 λ ) with λ K will act the same as the identity matrix. More precisely: Lemma 8.1. Let n 1. Then g GL n+1 (K) act trivially on P n (K) if and only if g is in K 1, i.e. g is a scalar multiple of the identity matrix. Proof. is clear. : By definition of P n (K), an element g acts trivially if for each v = (v 0, v 1,..., v n ) K n+1 there is a λ v K such that g v = λ v v. Take any two linearly independent vectors v and w in K n+1. Then λ v+ w ( v + w) = g ( v + w) = g v + g w = λ v v + λ w w shows that λ v+ w = λ v and λ v+ w = λ w. Therefore there is a λ K such that λ v = λ for all v. We define the projective linear group PGL n (K) as the quotient of GL n (K)/K. The class containing the matrix (a ij ) will be denoted by [a ij ]. Two matrices in GL n (K) represent the same element if they are a scalar multiple of each other, for instance [ ] [ ] a b λa λb = for all λ K c d λc λd By the previous lemma 8.1, the action of PGL n+1 (K) is now faithful on P n (K). 40

41 Proposition 8.2. If n > 2, then PGL n (K) acts 2-transitively on P n 1 (K), but not 3- transitively. If n = 2, then the action of PGL 2 (K) on P 1 (K) is sharply 3-transitive. Proof. Let n > 2. Given any two distinct points x and y in P n 1 (K), corresponding to two lines in K n passing through the origin, then it is easy to find a g GL n (K) sending x to any given line x and y to any given line y x. However if x, y, and z are three lines that belonging to the same plane in K n, then we can not send these three lines to three line that are not coplanar. Therefore the action is not 3-transitive. Now let n = 2. To show that the action is sharply 3-transitive, one has to show that the stabiliser of a point, say the point (0 : 1) at infinity, acts sharply 2-transitive on P 1 (K) \ {(1 : 0)} = K. This is done in an exercise. 8.1 Transvections Write V = K n. An element g GL n (K) is called a transvection if there is a hyperplane H V fixed point-wise by g and if g v v H for all v V. For instance the shear ( ) is a transvection with H = K e1. 2a a a H Figure 5: A picture of a transvection Proposition 8.3. If g is a transvection, there is a linear map ψ : V K and a vector a ker ψ such that g v = v + ψ( v) a. Conversely, given a non-trivial ψ : V K and a ker ψ, then t ψ, a ( v) = v + ψ( v) a is a transvection with fixed hyperplane H = ker ψ Proof. Let g be a transvection with fixed hyperplane H. Pick a u V \ H. Any v V can be written uniquely as v = w + λ u with w H and λ K. Define a linear map by ψ( v) = λ. Now we have g v = g w + λ g u = w + λ g u = v + λ ( u g u ). Setting a = g u u H = ker ψ proves the first part. The second point is straight forward to check. 41

42 For instance consider the matrix λ... 1 (8.1) equal to the identity except for a non-zero value λ in i-th row and the j-th column. This is a transvection t ψ, a with ψ( e k ) = { λ if k = j 0 otherwise where e i denotes the standard basis of K n. and a = e i H = {x j = 0}. Lemma 8.4. Let ϕ and ψ be two non-trivial linear maps V K. (a). t λψ, a = t ψ,λ a for all λ K and a ker ψ. (b). If t ψ, a = t ϕ, c, then there is a λ K such that ψ = λϕ and λ c = a. (c). g t ψ, a g 1 = t ψg 1,g a for all g GL n (K). (d). t ϕ, a t ψ, a = t ϕ+ψ, a for all a ker ψ ker ϕ. (e). t ψ, a t ψ, c = t ψ, a+ c for all a, c ker ψ. Proof. They are all straight forward calculations from the definition, apart from maybe (b). If t ψ, a = t ϕ, c, then ϕ( v) a = ψ( v) c for all v. In particular a and c are collinear, meaning that there is λ K such that λ c = a. Then λϕ( v) = ψ( v) for all v gives the rest. Lemma 8.5. Forany a K the matrix ( a 0 0 a 1 ) can be written as a product of transvections. Proof. ( ) ( ) ( ) ( ) ( ) ( ) a a 1 = a ( a 0 ) 0 a 1 Recall that SL n (K) is the subgroup of GL n (K) of matrices with determinant 1. Proposition 8.6. Let n 2. (a). All transvections belong to SL n (K). (b). If n 3, then any two non-trivial transvections are conjugate in SL n (K). (c). The centre of SL n (K) is { µ K µ n = 1 }. 42

43 (d). SL n (K) is generated by transvections. (e). If n 3, then [ SL n (K), SL n (K) ] = SL n (K). (f). If n = 2 and K 4, then [ SL 2 (K), SL 2 (K) ] = SL 2 (K). Proof. We start by proving that any two non-trivial transvections are conjugate in the group GL n (K). Let ψ and ϕ two non-trivial linear maps V K and let a ker ϕ and c ker ψ. Take a u outside ker ϕ and v outside ker ψ such that ϕ( u) = 1 and ψ( v) = 1. We want to define a linear g : V V. First we may pick any g which sends ker ϕ to ker ψ and which sends a to c in particular. If n 3, then the dimension of ker ϕ and ker ψ is larger than 2 and so we plenty of freedom to do so. Finally we can uniquely extend g to all V be imposing that u is sent to v. Then ϕ ( g 1 v ) = ϕ( u) = 1 = ψ( v) and ϕ g 1 and ψ have the same kernel. Hence ϕ g 1 = ψ. Therefore by item (c) in lemma 8.4, we have that g t ψ, a g 1 = t ϕ, c. (a). Since all transvections are conjugate in GL n (K) they all have the same determinant. The matrix in (8.1) has determinant 1. (b). As explained above, if n 3, then we still have some choice about g. We can choose it such that g has determinant 1. (c). Let z be an element of the centre of SL n (K). For all ψ : V K and a ker ψ, we must then have t ψ, a = z t ψ, a z 1 = t ψz 1,z a and by (b) of lemma 8.4, this means that there is a λ a K with z a = λ a a. As in the proof of lemma 8.1, this implies that z is a scalar multiple µ 1 of the identity matrix. However if z SL n (K), then 1 = det(z) = µ n. (d). Let g SL n (K). We want to multiply g with various transvections until we reach the identity matrix. This will show that every element in SL n (K) can be written as a product of transvections. Note that the transvections (8.1) correspond, when multiplied on the right of g, to the column operation of adding λ K times one column to another. When multiplied on the left it corresponds to a row operation. It is easy to see that with these operations any matrix in GL n (K) can be transformed to a diagonal matrix. Finally lemma 8.5 allows us to make all but one entry on the diagonal equal to 1. However, since the determinant did not change when multiplying with transvections, the result must now be the identity matrix. (e). Since all transvections are conjugate and they generate SL n (K), we only need to show that one transvection is among the commutators. Take any two non-trivial ϕ, ψ : V K and a ker ϕ ker ψ. Since there is a g SL n (K) such that g t ϕ, a g 1 = t ψ, a, we get g t ϕ, a g 1 t ϕ, a 1 = t ψ, a t ϕ, a = t ψ ϕ, a. 43

44 (f). The proof is a bit different than for n > 2 as not all transvections may be conjugate. However, ( any transvection will be conjugate in SL 2 (K) to a transvection of the form 1 ) 0 1, because we may bring the fixed line H to K e1 by conjugation with an element in SL 2 (K). Hence to prove that [SL 2 (K), SL 2 (K)] = SL 2 (K), we only need to be able to write all such matrices as commutators. We compute [ ( a 0 ) ( 0 a, 1 µ ) ] = ( )( a 0 1 µ 0 a )( a a )( 1 µ 0 1 ) = ( 1 µ(a 2 1) 0 1 for all a K and µ K. Now, if K has more than 3 elements, there is an a such that a 2 1 is non-zero. Fixing such an a, the appropriate choice of µ will allow us to write all the necessary matrices as commutators. ) We define PSL n (K) to be the image of SL n (K) in PGL n (K). It is equal to the quotient of SL n (K) by its centre. Theorem 8.7 (Jordan-Dickson). Let K be a finite field and let n 2. suppose that K 4. Then PSL n (K) is a simple group. If n = 2, we Proof. Let G = PSL n (K) and G = SL n (K). Then G acts faithfully and primitively on P n 1 (K): The action is faithful because we quotient out once again exactly by the kernel of the representation. Pick x P n 1 (K). The stabiliser H = Stab G (x) under the action of G is the intersection of G with the stabiliser under the action of PGL n (K). The latter is a maximal subgroup in PGL n (K). It is easy to show that H is maximal in G, too. Set H = Stab G(x). Further we define à to be the set of transvections t ψ, a with a x and ψ( a) = 0. Because t ψ, a t ϕ,λ a = t ψ, a t λϕ, a = t ψ+λϕ, a = t ϕ,λ a t ψ, a for all λ K and ψ( a) = ϕ( a) = 0, the set à is an abelian subgroup of G. Moreover, for all h H, we have h t ψ, a h 1 = t ψh 1,h a. now since x h a and (ψh 1 )(h a) = ψ( a) = 0, we obtain h t ψ, a h 1 à and therefore à H. By (d) in proposition 8.6, the elements in { gãg 1} generate all of G. g G Now set A to be the image of à in G. It is abelian and normal in H and { gag 1} g G generate all of G. Therefore Iwasawa s theorem 7.5 implies that any non-trivial normal subgroup of G contains G. However G = G under our assumptions. The two cases really need to be excluded as one can prove that PSL 2 (F 2 ) = S 3 and PSL 2 (F 3 ) = A 4 are both not simple. Here are all simple non-abelian groups of order smaller than 1000: Order Group 60 A 5 = PSL2 (F 4 ) = PSL 2 (F 5 ) PSL 2 (F 7 ) = PSL 3 (F 2 ) A 6 = PSL2 (F 9 ) 504 PSL 2 (F 8 ) 660 PSL 2 (F 11 ) 44

45 9 Finite rotation groups Let K = R and n 2. The orthogonal group is defined as { } O n = g GL n (R) gg t = 1 { = g GL n (R) (g v).(g w) = v. w v, w R n} { = g GL n (R) g v = v v R n} It is therefore the group of isometries of R n. It contains a normal subgroup of index 2, namely SO n consisting of all matrices in O n with determinant 1; these are the orientationreserving (or direct) isometries. For instance SO 2 is the group of all rotations around the origin. A reflection is an element g of order 2 in O n having a hyperplan H such that g fixes H point-wise and acts as multiplication by 1 on the orthogonal line to H. Our aim is to describe all finite subgroups of O n which is generated by reflections. 9.1 Two-dimensional case The dihedral group D n was defined as the isometry group of a regular n-gon in R 2. In particular, it is a finite subgroup of O 2. Let g 1 any reflections in D n, say whose axis passes through the vertex P of the n-gon. Let g 2 be the reflection whose axis passes through the centre of a edge ending at P. Then g 1 g 2 is a rotation by 2π n, so it has order n. Together Figure 6: The dihedral group is generated by 2 reflections g 1 and g 2 generate all of D n. We summarise this in a Coxeter diagram as I 2 (n) : n (9.1) where there should be n 2 edges between the two points. The two vertices of the graph represent the two generators g 1 and g 2. In general, the vertices are a chosen set of reflections that generate the group and we connect two vertices g and h if the order n of gh is larger than 2 by n 2 edges. Lemma 9.1. Any finite subgroup of O 2 generated by relections is a dihedral group D n for some integer n. 45

46 9.2 Three-dimensional examples Let G be the group of elements that fix the cube given by the vertices of the form (±1, ±1, ±1) R 3. This group acts on the set X of triples (V, E, F ) where V is a vertex of the cube, E is an edge of the cube ending on one side at V and F is a face of the cube having E as one of its sides. It is easy to see that G acts regularly on X; therefore G has as many elements as X, i.e. it has = 48 elements. Also G acts on the set Y consisting of the four body diagonals of the cube. This action is almost faithful, the only non-trivial element that acts trivially on Y is the central reflection j : (x, y, z) ( x, y, z). Hence there is a map G/ j S 4 which is injective. But both groups are of order 24, so this is an isomorphism. Moreover the subgroup G SO 3 maps isomorphically to S 4 under this map. Hence we have G = C 2 S 4, for j and any h G SO 3 commute. Proposition 9.2. The isometry group of the cube is generated by three reflections: g 1 : (x, y, z) (x, y, z) g 2 : (x, y, z) (y, x, z) g 3 : (x, y, z) (x, z, y) which satisfy (g 1 g 2 ) 4 = (g 2 g 3 ) 3 = (g 1 g 3 ) 2 = 1 Again we summarise this in the corresponding Coxeter graph B 3 : The proof is quite easy on the regular action of G on X. Next, we consider the group of the octahedron. Because the centre of the faces of an octahedron form a cube and the centre of the faces of a cube form an octahedron, we say that the two regular polyhedrons are dual to each other. See figure 7. It is clear that they have the same group of isometries. Figure 7: The cube and the octahedron are dual The dodecahedron and the icosahedron are also dual to each other. Their group of isometries is generated by 3 elements g 1, g 2, g 3 with the relations (g 1 g 2 ) 5 = (g 2 g 3 ) 3 = (gg 1 g 3 ) 2 = 1, so the corresponding Coxeter graph is H 3 : (9.2) 46

47 One can show that the group is isomorphic to C 2 A 5. The faces of the icosahedron can be coloured with 5 colours in such a say as the action of the group on the colours identifies G SO 3 with A 5. See figure 8 This is also the isometry group of the football, which can Figure 8: The icosahedron and its colouring with 5 colours be obtained by cutting off bits at each vertex from a icosahedron. Finally the tetrahedron, which is dual to itself, has the Coxeter graph A 3 : (9.3) It simply identifies with the group S 4 viewed as the permutations of the vertices of the tetrahedron. See figure 9 Figure 9: The tetrahedron 9.3 Infinite families There are three families of finite groups of reflections in n-space R n. First, the group S n can be viewed as acting by permutations on the standard basis of R n. This group is generated by the n 1 transpositions g 1 = (1 2), g 2 = (2 3),..., g n 1 = (n 1 n) and the Coxeter graph is A n 1 : (9.4) as we have (g i g j ) 2 = 1 if j = i + 1 and (g i g j ) 3 = 1 if j > i + 1. This group is the group of isometries of the so-called (n 1)-simplex, whose vertices are the n end-points of the basis vectors. It is a regular polytope in the n 1-dimensional affine subspace x 1 + x x n = 0. It generalises the triangle and the tetrahedron. Next, we have the group C2 n S n where the action of S n on C2 n = F n 2 is by permutation of a basis. We can view this as the group of permutations of the standard basis in 47