13 Solutions for Section 6

13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution The group S 3 contains the identity permutation, id, three transpositions (1 2), (1 3), (2 3) and two 3-cycles (1 2 3) and (1 3 2). id (1 2) (1 3) (2 3) (1 2 3) (1 3 2) id id (1 2) (1 3) (2 3) (1 2 3) (1 3 2) (1 2) (1 2) id (1 3 2) (1 2 3) (2 3) (1 3) (1 3) (1 3) (1 2 3) id (1 3 2) (1 2) (2 3) (2 3) (2 3) (1 3 2) (1 2 3) id (1 3) (1 2) (1 2 3) (1 2 3) (1 3) (2 3) (1 2) (1 3 2) id (1 3 2) (1 3 2) (2 3) (1 2) (1 3) id (1 2 3) Exercise 6.3 Give an example of a group G, a subgroup H and an element a G such that the right and left cosets ah and Ha are not equal. Solution (Notice that there is no point looking in abelian groups since if the multiplication is commutative then right and left cosets coincide.) The simplest non-abelian group is S 3 so that s a good place to start looking. It will follow from what we do later that it s no good taking a normal subgroup since, by Lemma 8.2, the right and left cosets would coincide. By Example 8.5 a subgroup of index 2 must be normal, so that rules out using the subgroup of S 3 consisting of the 3-cycles and the identity. All this leads to trying the following example, which works.) Let G = S 3, let H = {id,(1 2)} be the (2-element cyclic) subgroup generated by (1 2). Take a = (1 2 3). Then ah = {(1 2 3)id,(1 2 3)(1 2)} = {(1 2 3),(1 3)}. On the other hand Ha = {id(1 2 3),(1 2)(1 2 3)} = {(1 2 3),(2 3)} ah, as required. Exercise 6.6 Write down the group tables, where the operation is addition, for Z 4 and also for Z 5. Do the same for the field F 4 with 4 elements (see Section 6). Solution I think that computing the tables for Z 4 and Z 5 should present no problems. And the table for F 4 can be found in Example 1.3 of the notes (the third example is F 4 ). Exercise 6.8 Find U(R) and draw up the group table for each of the following rings R: Z; Z 5 ; Z 8 ; F 4. Solution The only invertible elements of Z are ±1 so the table is just the following. 1 1 1 1 1 1 1 1 The invertible elements of Z 5 are the classes of 1, 2, 3, 4 so, writing just a for [a] 5, we have the following. 69

1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 The invertible elements of Z 8 are the classes of 1, 3, 5, 7 so the table is the following. 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 The invertible elements of F 4 are 1, α, 1+α where α is a root of X 2 +X +1, so the table is the following. 1 α 1 + α 1 1 α 1 + α α α 1 + α 1 1 + α 1 + α 1 α Exercise 6.9 Let R be the ring M 2 (Z) of 2 2 matrices with integer entries. Find some elements in the group of units of R and ( compute ) their orders. Now replace Z by R. What is a criterion for U(M 2 (R)? Given n can you find an element of order n in U(M 2 (R))? 1 0 0 1 Solution A couple of elements of order 2 are and. The 0 1 1 0 0 1 1 1 matrix has order 4. The matrix has infinite order. 1 0 0 1 When we move to matrices with entries from R then we have ( the criterion ) from linear algebra (for invertibility of a matrix over a field): is invertible iff det 0, that is, iff ad bc 0. For the last part, think geometrically: a 2 2 matrix with real entries can be thought of as a linear transformation of the plane and, conversely every linear transformation of the plane can be represented (with respect to any given basis) as a 2 2 matrix. Rotation about the origin ( by 2π/n is a linear transformation ) cos(2π/n) sin(2π/n) of order n, so the corresponding matrix is of order sin(2π/n) cos(2π/n) n (you can check this directly if you want, using trigonometric identities, but that s a more complicated argument than the one using the link to geometry). Exercise 6.11 Denote by D n the group of symmetries of a regular n-gon. Show that the dihedral group D n has 2n elements, of which n are rotations and n are reflections. Draw up the group table for D 3. Solution The n rotations and n reflections are all, I imagine, fairly obvious. To argue that there are no more symmetries than this: choose one vertex; there are n choices of where a symmetry can send it; having made that choice there are only two further choices - whether to keep its two immediate neighbours in whatever order they are or to switch them - so n 2 choices in all. 70

For the table of D 3 let ρ denote rotation by 2π/3 and let σ be any reflection. Then the elements of D 3 are: id, ρ, ρ 2, σ, σρ, σρ 2 and the table is as follows. id ρ ρ 2 σ σρ σρ 2 id id ρ ρ 2 σ σρ σρ 2 ρ ρ ρ 2 id σρ 2 σ σρ ρ 2 ρ 2 id ρ σρ σρ 2 σ σ σ σρ σρ 2 id ρ ρ 2 σρ σρ σρ 2 σ ρ 2 id ρ σρ 2 σρ 2 σ σρ ρ ρ 2 id If you choose to write the elements of D 3 in the forms id, ρ, ρ 2, σ, ρσ, ρ 2 σ then you will get a similar table. Exercise 6.13 Show that the symmetric group S 3 and the dihedral group D 3 are isomorphic by bringing their group tables to the same form. Define an isomorphism from S 3 to D 3. Now define another isomorphism from S 3 to D 3. Solution The group table for S 3 is given in the solution for Exercise 7.2, that for D 3 is given in the solution for Exercise 7.14 above and clearly they re not in the same form. But we can try re-arranging the order of elements in, say S 3. It s clear that an isomorphism will take an element of order 3 to an element of order 3 so let s list the elements of S 3 in the order id, (1 2 3), (1 3 2) = (1 2 3) 2, then we ll choose one of the transpositions, say (1 2), to go next. So far, we ve matched up ρ with (1 2 3), hence ρ 2 with (1 3 2), and we ve matched σ with (1 2), so, if we follow the order in the table for D 3 given above we should match the next in the list, σρ, with (1 2)(1 2 3) = (2 3), so that leaves (1 3) for the last on the list. So here s a re-arranged group table for S 3. Note that it does have exactly the same form as the table for D 3 above, which shows that the groups are isomorphic. id (1 2 3) (1 3 2) (1 2) (2 3) (1 3) id id (1 2 3) (1 3 2) (1 2) (2 3) (1 3) (1 2 3) (1 2 3) (1 3 2) id (1 3) (1 2) (2 3) (1 3 2) (1 3 2) id (1 2 3) (2 3) (1 3) (1 2) (1 2) (1 2) (2 3) (1 3) id (1 2 3) (1 3 2) (2 3) (2 3) (1 3) (1 2) (1 3 2) id (1 2 3) (1 3) (1 3) (1 2) (2 3) (1 2 3) (1 3 2) id The matching (or pairing) that we referred to above is the description of an isomorphism, θ say, namely that defined by θ(ρ) = (1 frm e 3), θ(ρ 2 ) = (1 3 2), θ(σ) = (1 2), θ(σρ) = (2 3), θ(σρ 2 ) = (1 3). You could get another isomorphism by either sending ρ to the other element, (1 3 2), of order 3 and/or sending σ to either of the other elements, (1 3) and (2 3), of order 2. Exercise 6.14 Check that G H is a group. Prove that G H H G (note that, to do this, you should produce an isomorphism θ). Proof Identity element of G H is (e, e) = (e G, e H ) since, given (g,h) G H we have (g,h)(e, e) = (ge, he) = (g,h) = (eg, eh) = (e, e)(g,h). Given (g, h) G H we claim that (g 1, h 1 ) is its inverse: for (g,h)(g 1, h 1 ) = (gg 1, hh 1 ) = (e, e) = (g 1 g,h 1 h) = (g 1, h 1 )(g, h). Finally we have to check associativity but this is also very easy and left for you to write out. An isomorphism between G H and H G is given by switching the coordinates: define θ : G H H G by θ(g,h) = (h, g). This is ijec- 71

tion: it is an injective since, if θ(g,h) = θ(g, h ) then (h, g) = (h, g ) so h = h and g = g so (g,h) = (g, h ), and it is a surjection since, given an element a of H G this element has the form (h, g) for some h H and g G so a = θ(g, h). Finally we check that θ is a homomorphism: we have θ ( (g, h)(g, h ) ) = θ(gg, hh ) = (hh, gg ) = (h, g)(h, g ) = θ(g,h)θ(g, h ), as required. Exercise 6.15 Draw up the multiplication table for the group of symmetries of a rectangle which is not a square. Show that this group is isomorphic to Z 2 Z 2 (where addition is the operation). Solution Let σ denote one reflection in a midline and let τ denote the other (or, if you prefer, take τ to be the rotation through an angle of π). Then the table is as follows. id σ τ στ id id σ τ στ σ σ id στ τ τ τ στ id σ στ στ τ σ id The table for Z 2 Z 2 is as follows. + (0, 0) (0, 1) (1, 0) (1, 1) (0,0) (0,0) (0,1) (1,0) (1,1) (0,1) (0,1) (0,0) (1,1) (1,0) (1,0) (1,0) (1,1) (0,0) (0,1) (1, 1) (1, 1) (1, 0) (0, 1) id Exercise 6.16 Show that the group of rotations of a cube is isomorphic to S 4. How many symmetries of a cube are there? Show, by considering the relation between an octahedron and a cube, that an octahedron has the same group of symmetries. Solution The simplest way of doing the first is to look at the 4 body-diagonals of the cube, note that any rotation of the cube is determined by its action on these diagonals (i.e. can be thought of as a permutation of these four lines) and then check that, conversely, every permutation of these 4 lines can be induced by a rotation of the cube. Notice that if you keep all these diagonals fixed but interchange the two vertices at the opposite ends of one of them (achievable by a reflection) then you have a symmetry which is not achievable by a rotation. So there are also those obtained by composing a rotation with a reflection; these form a single coset of the subgroup of rotations within the group of symmetries (the fact that they all belong to the same coset can, for instance be proved by considering determinants, necessarily ±1, of the linear maps corresponding to symmetries). This gives 48 symmetries of the cube in total. Put a vertex at the centre of each face of the cube and join the vertices to get an octahedron. Put a vertex at the centre of each face of the octahedron and join the vertices to get a cube (a smaller version of the first one, sitting inside it). So any symmetry of the cube gives a symmetry of the octahedron and any symmetry of the octahedron gives a symmetry of the small, hence the original, cube. Exercise 6.25 Draw up some of the group table for A 4. 72

Solution Since there are 12 elements in A 4 the whole table would be a 12 12. An exercise is useful only so long as you are gaining something from it and I expect that no-one would gain much from ploughing on after a moderate amount of the table has been filled in. I will just list the 12 elements of A 4 : id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3). You should do a few of the multiplications as illustrations of the fact that this set is closed under multiplication. Check your computations with someone else if you have any doubts over them. Exercise 6.26 Show that the group of symmetries of a tetrahedron is S 4. Show that the subgroup of those symmetries which are realisable as rotations (i.e. without going into a mirror 3-space) is isomorphic to A 4. (In each case, use reasoning rather than try to draw up the (rather large) group table.) Solution The first part will have been done in lectures. One argument for the second is first to observe that any rotation can be obtained by combining rotations which fix a vertex and rotate the other three vertices - this corresponds to a products of 3-cycles, hence is even. All other symmetries can be obtained by a single reflection in a mirror, say that corresponding to the (odd) transposition (2 3), followed by a rotation (hence an even permutation) in the mirror world. So the group of all symmetries falls into two cosets of the subgroup consisting of those which are (combinations of) rotations, so it follows that this subgroup corresponds to A 4. Exercise 6.28 Show that, in the above action of G on itself, the orbit of each element is the whole group and the stabiliser of each element is the trivial subgroup. Proof (The action being referred to is the natural action of G on itself: G G G given by (g, h) gh.) Let h, h G. Then h h 1 h = h, which shows that h is in the orbit of h. If g G stabilises h then gh = h so g = e so Stab(h) = {e} for every h H. Exercise 6.30 Compute some orbits and stabilisers in geometric examples and also for the action of the symmetric group S n on {1,...,n}. For example, compute the stabiliser, in the group of symmetries of a cube, of a vertex of the cube. Also consider the action of this group of symmetries on the set of edges of the cube and compute the stabiliser of an edge. Compute the orbit, under the group of symmetries, of each vertex of a square-based regular (as far as it can be) pyramid. Compute the stabiliser of 4 under the natural action of S 4 on {1,2,3,4}. Compute the stabiliser of the set {1,2} under the action of S 5 on two-element subsets of {1,...,5} which is induced by the action of S 5 on {1,...,5}. Comments re Solution For example, the stabiliser of a vertex of a cube is S 3 (rotate and reflect the 3 neighbours of that vertex). The stabiliser of an edge is a 2-element group, i.e. generated by an appropriate transposition (all that can be done is reflect in the mid-point of the edge). In the square-based pyramid, the top vertex is fixed by all symmetries so is in an orbit with just that one element; the four base vertices lie in a single 4-element orbit. The permutations of S 4 which fix 4 are the 6 permutations which essentially consist of a permutation in S 3 expanded by fixing 4 to a permutation in S 4. In the last example the action of S 5 on two-element sets {a, b} with a b, is: given 73

σ S 5 set σ{a, b} = {σ(a), σ(b)}. Then σ{1,2} = {1,2} iff σ either fixes or interchanges 1 and 2. So permutations of the form τρ where ρ is a permutation of 3,4,5 and σ is either the identity permutation or the transposition (1 2) (in particular there are 12 elements in the stabiliser of {1,2}. Exercise 6.32 Check the Orbit-Stabiliser Theorem in the examples you looked at above (that stabilisers are subgroups and that the index of the stabiliser equals the number of elements in the orbit). Solution (partial) For example, the stabiliser of 4 has six elements, they form a group, and the orbit of 4 is clearly {1,2,3,4} so has 4 = 24/6 elements, as predicted by the Orbit-Stabiliser Theorem. Exercise 6.34 Show that being conjugate is an equivalence relation on the set of subgroups of G. Proof Let H be a subgroup of G; then e 1 He = H so H is conjugate to itself (reflexivity). Suppose that H,K are subgroups with K = a 1 Ha; then H = aka 1 = (a 1 ) 1 K(a 1 ) so we have symmetry. Finally suppose that H,K, L are subgroups with K = a 1 Ha and L = b 1 Kb; then L = b 1 a 1 Hab = (ab) 1 H(ab) so L is conjugate to H (transitivity). Exercise 6.39 Consider the natural action of S 3 on K[X 1, X 2, X 3 ]. Compute the orbit and stabiliser of each of the following polynomials: X 1, X 1 + X 2, X 1 + X 2 + X 3, X 1 X 2 3 + X 3 X 2 1, X 1 X 2 3 + X 3 X 2 1, X 1 X 2 2 + X 2 X 2 3 + X 3 X 2 1, X 1 + X 2 2 + X 3 3. Solution X 1 : {X 1, X 2, X 3 } (orbit); {id,(2 3)} (stabiliser) X 1 + X 2 : {X 1 + X 2, X 1 + X 3, X 2 + X 3 } (orbit); {id,(1 2)} (stabiliser) X 1 + X 2 + X 3 : {X 1 + X 2 + X 3 } (orbit); S 3 (stabiliser) (this polynomial is invariant under the action of S 3 - it is symmetric in X 1, X 2 and X 3 ) X 1 X 2 3+X 3 X 2 1: {X 1 X 2 3+X 3 X 2 1, X 1 X 2 2+X 2 X 2 1, X 2 X 2 3+X 3 X 2 2 } (orbit); {id,(1 3)} (stabiliser) X 1 X 2 2 +X 2 X 2 3 +X 3 X 2 1: {X 1 X 2 2 +X 2 X 2 3 +X 3 X 2 1, X 1 X 2 3 +X 3 X 2 2 +X 2 X 2 1 } (orbit); {id,(1 2 3),(1 3 2)} (stabiliser) X 1 + X 2 2 + X 3 3: {X 1 + X 2 2 + X 3 3, X 1 + X 2 3 + X 3 2, X 2 + X 2 1 + X 3 3, X 2 + X 2 3 + X 3 1, X 3 + X 2 1 + X 3 2, X 3 + X 2 2 + X 3 1 } (orbit); {id} (stabiliser) Note that, in accordance with the Orbit-Stabiliser Theorem, in each case the number of elements in the orbit times the number of elements in the stabiliser equals 6, the number of elements in the whole group S 3. (Indeed you might have found this useful in answering, in the sense that you can stop searching when the product reaches 6.) Exercise 6.40 Let R = M 2 (Z) be the ring of 2 2 matrices with entries from the ring Z of integers. Determine the condition on a matrix to be in the group of units of this ring and give some examples of units. Solution The first observation to make is that if is invertible then a b it must have determinant ±1: for, if c d = 1 then, taking determinants, we have (ad bc)(a d b c ) = 1 which, since we re dealing with integers, means that both determinants are ±1 (if we were dealing with M 2 (Q) 74

then it would all be much simpler since a matrix over any field is invertible iff its determinant is non-zero). Next, recalling how to write down the inverse of a 2 2 matrix, if d b has determinant ±1 then it is invertible, with inverse (ad bc) 1. c a So the condition for invertibility is having determinant ±1. 1 0 So this gives immediate examples of invertible elements like, 0 1 1 53 0 1, but you can find less obvious examples by solving 0 1 1 72 the equation ad bc = 1. Think of fixing d and c to be two coprime integers, then you know that there will be (lots of) integers a, b such that ad bc = 1 (or = 1). For instance, choose d = 3, c = 4; ( then (1)3 ) (1)(4) ( = 1 and ) 1 1 5 4 ( 5)3 ( 4)4 = 1 give the invertible matrices and, 4 3 4 3 3 1 3 4 with inverses and respectively. Obviously, there 4 1 4 5 are lots more. Exercise 6.41 Let R = M 2 (Z 4 ) be the ring of 2 2 matrices with entries from the ring Z 4 of integers modulo 4. Find some elements of the group of units of this ring. Is this group U(R) of units of R abelian? What is a criterion for a matrix in M 2 (Z 8 ) to be invertible? Solution We can begin as above: assuming that a matrix is invertible it follows that its determinant must be an invertible element of Z 4. And conversely, if it determinant is invertible then the usual formula for the inverse of a 2 2 matrix shows that its inverse also is in M 2 (Z 4 ). The only invertible elements of Z 4 are ±1 (i.e. 1 and 3) so criterion for invertibility is having determinant 1 or 3. Here are two invertible ( elements) which( don t commute ) (hence which show 0 1 1 2 that U(R) is not abelian): and. 1 0 3 0 In the case of M 2 (Z 8 ) the criterion for invertibility is that the determinant should be invertible in Z 8, that is, the determinant should be 1, 3, 5 or 7. Exercise 6.42 Let R = M 2 (F 4 ) be the ring of 2 2 matrices with entries from the field F 4 with 4 elements. Find some elements of the group of units of this ring. How many elements are in the group U(R) of units of R? How many of these have determinant 1? Solution Since F 4 is a field a matrix is invertible iff its determinant is nonzero. In writing down some invertible matrices you need to give yourself some notation. Remember that F 4 is the extension of Z 2 by an element which is a root of the (only) irreducible quadratic, X 2 +X +1, so you can take the elements of F 4 to be 0, 1, ( α and 1+α where ) α satisfies α 2 +α+1 = 0, that is α 2 = α+1. So, α 1 for instance, is not invertible since its determinant is α + α 1 1 + α 2 1 α 0 which equals 0. On the other hand the determinant of is α + α 1 1 + α 2 75

which equals 1, so this matrix is invertible. Counting the invertible matrices is more interesting. First note that there are 4 4 = 256 matrices in M 2 (Z 4 ). In constructing an invertible matrix, the first row can have any entries except 0 0 - that makes 4 2 1 = 15 possibilities for the first row. The second row can be any 2-vector which is linearly independent from the first row (think of the rows as vectors in 2-dimensional space over F 4 ). There are 4 2 = 16 2-vectors altogether and we have to avoid the multiples of the first row (including the zero vector). That means 4 vectors to be avoided, but any of the other 12 will do for the second row. Therefore there are 15 12 = 180 invertible matrices (and hence 76 matrices with determinant 0). To do the last part, use that the determinant map can be regarded as a map from the group of invertible matrices (under multiplication) to the group of non-zero elements, {1, α,1 + α}, of elements of F 4 (under multiplication). The kernel of this map is the set of matrices with determinant 1 and the other two cosets of the kernel are: the set of matrices with determinant α and the set of matrices with determinant 1 + α. By basic group theory all these three sets have the same number of elements, 60 each. Therefore there are 60 matrices with determinant 1. Exercise 6.43 Identify the groups of symmetries of the following geometric figures: (i) a triangle; (ii) a square; (iii) a rectangle which is not a square; (iv) a hexagon; (v) a tetrahedron; (vi) a cube. Why is the group of symmetries of a cube the same as that of a regular octahedron? (Or, for that matter, why is the group of symmetries of a dodecahedron the same as that of an icosahedron?) Solution (i) D 3 S(3) (all permutations of the vertices can be realised by symmetries); (ii) D 4 (i.e. just 8 symmetries; in particular not all of the 24 permutations of the vertices can be realised); (iii) Z 2 Z 2 (the Klein 4-group); (iv) D 6 (six rotations, six reflections); (v) S(4) (fixing one vertex, all six symmetries of the triangle formed by the other three can be realised by a rotation or a reflection, but you can also move that vertex to any of the other four positions); (vi) S(4) (look at the 4 body diagonals: each symmetry of the cube is determined by how it permutes these line segments and, on the other hand, every permutation of them is realised by a symmetry of the cube. So the symmetry group of the cube is isomorphic to the symmetry group of these four objects, that is, to S(4)). Draw the vertex in the middle of each of the 6 faces of the cube and join them to make a regular octahedron. Any symmetry acts equally as a symmetry of the cube or of this octahedron (and note that you can go in the same way from a regular octahedron to a cube; and similarly the dodecahedron and icosahedron are paired). Exercise 6.44 Let X be a square-based pyramid with all sides of equal length. Find the group G of symmetries of X and compute the stabiliser of each vertex. Solution The apex is a vertex of valency 4 and the others have valency 3 so the apex is fixed by every symmetry. So, in effect, the group of symmetries is that of the square base, that is D 4 (the dihedral group with 8 elements). The stabiliser of the apex is the whole group D 4. If a symmetry fixes ase vertex then it must either fix both immediate neighbours (so then must be the identity symmetry) or it switches the two neighbours (and fixes the diagonally opposite vertex). The second symmetry has order 2 and we see that the stabiliser of a vertex is, up to isomorphism, the cyclic group of order 2. More precisely, diagonally opposite 76

vertices on the base have the same stabilisers (consisting of the identity and one of the two diagonal reflections). Even more precisely, if we label the base vertices in order 1 (joined to) 2 (joined to) 3 (joined to) 4 then the stabiliser of 1 is the identity together with the symmetry given by the permutation (2 4), and this is also the stabiliser of 3. And the stabiliser of 2 (=the stabiliser of 4) is the identity and the permutation given by (1 3). The symmetry given by σ = (1 2 3 4) takes vertex 1 to vertex 2, hence conjugation by it induces an isomorphism between the respective stabilisers - we check explicitly: stab(2) = σstab(1)σ 1 = (1 2 3 4)(2 4)(4 3 2 1) = (1 3). 77