13 Solutions for Section 6


 Jack Ryan
 2 years ago
 Views:
Transcription
1 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 3cycles (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 nonabelian 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 3cycles and the identity. All this leads to trying the following example, which works.) Let G = S 3, let H = {id,(1 2)} be the (2element 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 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
2 The invertible elements of Z 8 are the classes of 1, 3, 5, 7 so the table is the following 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 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))? Solution A couple of elements of order 2 are and. The matrix has order 4. The matrix has infinite order 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 ngon. 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
3 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 rearranging 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 rearranged 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
4 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 bodydiagonals 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
5 Solution Since there are 12 elements in A 4 the whole table would be a An exercise is useful only so long as you are gaining something from it and I expect that noone 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 3space) 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 3cycles, 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 squarebased 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 twoelement 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 2element group, i.e. generated by an appropriate transposition (all that can be done is reflect in the midpoint of the edge). In the squarebased 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 4element 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 twoelement sets {a, b} with a b, is: given 73
6 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 OrbitStabiliser 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 OrbitStabiliser 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 X 3 X 2 1, X 1 X X 3 X 2 1, X 1 X X 2 X X 3 X 2 1, X 1 + X 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 X 3 3: {X 1 + X X 3 3, X 1 + X X 3 2, X 2 + X X 3 3, X 2 + X X 3 1, X 3 + X X 3 2, X 3 + X X 3 1 } (orbit); {id} (stabiliser) Note that, in accordance with the OrbitStabiliser 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
7 then it would all be much simpler since a matrix over any field is invertible iff its determinant is nonzero). 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 ± So this gives immediate examples of invertible elements like, , but you can find less obvious examples by solving 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 ) ( 5)3 ( 4)4 = 1 give the invertible matrices and, with inverses and respectively. Obviously, there 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 that U(R) is not abelian): and 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 α 2 1 α 0 which equals 0. On the other hand the determinant of is α + α α 2 75
8 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 that makes = 15 possibilities for the first row. The second row can be any 2vector which is linearly independent from the first row (think of the rows as vectors in 2dimensional space over F 4 ). There are 4 2 = 16 2vectors 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 = 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 nonzero 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 4group); (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 squarebased 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
9 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 = ( ) 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 = ( )(2 4)( ) = (1 3). 77
DMATH Algebra I HS 2013 Prof. Brent Doran. Solution 5
DMATH Algebra I HS 2013 Prof. Brent Doran Solution 5 Dihedral groups, permutation groups, discrete subgroups of M 2, group actions 1. Write an explicit embedding of the dihedral group D n into the symmetric
More information1 Symmetries of regular polyhedra
1230, notes 5 1 Symmetries of regular polyhedra Symmetry groups Recall: Group axioms: Suppose that (G, ) is a group and a, b, c are elements of G. Then (i) a b G (ii) (a b) c = a (b c) (iii) There is an
More informationCLASSIFYING FINITE SUBGROUPS OF SO 3
CLASSIFYING FINITE SUBGROUPS OF SO 3 HANNAH MARK Abstract. The goal of this paper is to prove that all finite subgroups of SO 3 are isomorphic to either a cyclic group, a dihedral group, or the rotational
More informationChapter 7. Permutation Groups
Chapter 7 Permutation Groups () We started the study of groups by considering planar isometries In the previous chapter, we learnt that finite groups of planar isometries can only be cyclic or dihedral
More informationGroup Theory (MA343): Lecture Notes Semester I Dr Rachel Quinlan School of Mathematics, Statistics and Applied Mathematics, NUI Galway
Group Theory (MA343): Lecture Notes Semester I 20132014 Dr Rachel Quinlan School of Mathematics, Statistics and Applied Mathematics, NUI Galway November 21, 2013 Contents 1 What is a group? 2 1.1 Examples...........................................
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationConstruction and Properties of the Icosahedron
Course Project (Introduction to Reflection Groups) Construction and Properties of the Icosahedron Shreejit Bandyopadhyay April 19, 2013 Abstract The icosahedron is one of the most important platonic solids
More informationGeometric Group Theory Homework 1 Solutions. e a b e e a b a a b e b b e a
Geometric Group Theory Homework 1 Solutions Find all groups G with G = 3 using multiplications tables and the group axioms. A multiplication table for a group must be a Latin square: each element appears
More informationNotes on finite group theory. Peter J. Cameron
Notes on finite group theory Peter J. Cameron October 2013 2 Preface Group theory is a central part of modern mathematics. Its origins lie in geometry (where groups describe in a very detailed way the
More informationEach object can be assigned to a point group according to its symmetry elements:
3. Symmetry and Group Theory READING: Chapter 6 3.. Point groups Each object can be assigned to a point group according to its symmetry elements: Source: Shriver & Atkins, Inorganic Chemistry, 3 rd Edition.
More informationProblem Set 1 Solutions Math 109
Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twentyfour symmetries if reflections and products of reflections are allowed. Identify a symmetry which is
More informationENUMERATION BY ALGEBRAIC COMBINATORICS. 1. Introduction
ENUMERATION BY ALGEBRAIC COMBINATORICS CAROLYN ATWOOD Abstract. Pólya s theorem can be used to enumerate objects under permutation groups. Using group theory, combinatorics, and many examples, Burnside
More information3. QUADRATIC CONGRUENCES
3. QUADRATIC CONGRUENCES 3.1. Quadratics Over a Finite Field We re all familiar with the quadratic equation in the context of real or complex numbers. The formula for the solutions to ax + bx + c = 0 (where
More informationGROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS
GROUP ACTIONS ON SETS WITH APPLICATIONS TO FINITE GROUPS NOTES OF LECTURES GIVEN AT THE UNIVERSITY OF MYSORE ON 29 JULY, 01 AUG, 02 AUG, 2012 K. N. RAGHAVAN Abstract. The notion of the action of a group
More informationGroup Theory G13GTH. Chris Wuthrich. 1 Definitions and examples Examples Direct and semidirect products Small groups...
Group Theory G13GTH Chris Wuthrich 2012 13 Contents 1 Definitions and examples 3 1.1 Examples..................................... 3 1.2 Direct and semidirect products......................... 4 1.3 Small
More information3 Congruence arithmetic
3 Congruence arithmetic 3.1 Congruence mod n As we said before, one of the most basic tasks in number theory is to factor a number a. How do we do this? We start with smaller numbers and see if they divide
More informationG = G 0 > G 1 > > G k = {e}
Proposition 49. 1. A group G is nilpotent if and only if G appears as an element of its upper central series. 2. If G is nilpotent, then the upper central series and the lower central series have the same
More information7. The GaussBonnet theorem
7. The GaussBonnet theorem 7. Hyperbolic polygons In Euclidean geometry, an nsided polygon is a subset of the Euclidean plane bounded by n straight lines. Thus the edges of a Euclidean polygon are formed
More informationChapter 7: Products and quotients
Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products
More informationThe noblest pleasure is the joy of understanding. (Leonardo da Vinci)
Chapter 6 Back to Geometry The noblest pleasure is the joy of understanding. (Leonardo da Vinci) At the beginning of these lectures, we studied planar isometries, and symmetries. We then learnt the notion
More informationThe determinant of a skewsymmetric matrix is a square. This can be seen in small cases by direct calculation: 0 a. 12 a. a 13 a 24 a 14 a 23 a 14
4 Symplectic groups In this and the next two sections, we begin the study of the groups preserving reflexive sesquilinear forms or quadratic forms. We begin with the symplectic groups, associated with
More informationAnother new approach to the small Ree groups
Another new approach to the small Ree groups Robert A. Wilson Abstract. A new elementary construction of the small Ree groups is described. Mathematics Subject Classification (2000). 20D06. Keywords. Groups
More informationThe Mathematics of Origami
The Mathematics of Origami Sheri Yin June 3, 2009 1 Contents 1 Introduction 3 2 Some Basics in Abstract Algebra 4 2.1 Groups................................. 4 2.2 Ring..................................
More informationElements of Abstract Group Theory
Chapter 2 Elements of Abstract Group Theory Mathematics is a game played according to certain simple rules with meaningless marks on paper. David Hilbert The importance of symmetry in physics, and for
More informationEquivalence relations
Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence
More informationLecture #5 (Dihedral group, Symmetric group, The Quaternion group, Klein4 group) MTH 232, Abstract Algebra
Lecture #5 (Dihedral group, Symmetric group, The Quaternion group, Klein4 group) MTH 232, Abstract Algebra February 20, 2012 Dihedral groups An important family of examples of groups is the class of groups
More informationAbstract Algebra Cheat Sheet
Abstract Algebra Cheat Sheet 16 December 2002 By Brendan Kidwell, based on Dr. Ward Heilman s notes for his Abstract Algebra class. Notes: Where applicable, page numbers are listed in parentheses at the
More informationGROUPS ACTING ON A SET
GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for
More information4. FIRST STEPS IN THE THEORY 4.1. A
4. FIRST STEPS IN THE THEORY 4.1. A Catalogue of All Groups: The Impossible Dream The fundamental problem of group theory is to systematically explore the landscape and to chart what lies out there. We
More informationCONJUGATION IN A GROUP KEITH CONRAD
CONJUGATION IN A GROUP KEITH CONRAD 1. Introduction A reflection across one line in the plane is, geometrically, just like a reflection across any other line. That is, any two reflections in the plane
More informationGroup Theory. Contents
Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation
More information3. Some commutative algebra Definition 3.1. Let R be a ring. We say that R is graded, if there is a direct sum decomposition, M n, R n,
3. Some commutative algebra Definition 3.1. Let be a ring. We say that is graded, if there is a direct sum decomposition, = n N n, where each n is an additive subgroup of, such that d e d+e. The elements
More informationGROUP ACTIONS KEITH CONRAD
GROUP ACTIONS KEITH CONRAD 1. Introduction The symmetric groups S n, alternating groups A n, and (for n 3) dihedral groups D n behave, by their very definition, as permutations on certain sets. The groups
More informationComputing the Symmetry Groups of the Platonic Solids With the Help of Maple
Computing the Symmetry Groups of the Platonic Solids With the Help of Maple Patrick J. Morandi Department of Mathematical Sciences New Mexico State University Las Cruces NM 88003 USA pmorandi@nmsu.edu
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationMathematics Course 111: Algebra I Part II: Groups
Mathematics Course 111: Algebra I Part II: Groups D. R. Wilkins Academic Year 19967 6 Groups A binary operation on a set G associates to elements x and y of G a third element x y of G. For example, addition
More informationGeometric Transformations
Geometric Transformations Definitions Def: f is a mapping (function) of a set A into a set B if for every element a of A there exists a unique element b of B that is paired with a; this pairing is denoted
More informationBinary Operations. Discussion. Power Set Operation. Binary Operation Properties. Whole Number Subsets. Let be a binary operation defined on the set A.
Binary Operations Let S be any given set. A binary operation on S is a correspondence that associates with each ordered pair (a, b) of elements of S a uniquely determined element a b = c where c S Discussion
More informationProof. Right multiplication of a permutation by a transposition of neighbors either creates a new inversion or kills an existing one.
GROUPS AROUND US Pavel Etingof Introduction These are notes of a minicourse of group theory for high school students that I gave in the Summer of 2009. This minicourse covers the most basic parts of
More information1.5 Elementary Matrices and a Method for Finding the Inverse
.5 Elementary Matrices and a Method for Finding the Inverse Definition A n n matrix is called an elementary matrix if it can be obtained from I n by performing a single elementary row operation Reminder:
More informationTo introduce the concept of a mathematical structure called an algebraic group. To illustrate group concepts, we introduce cyclic and dihedral groups.
1 Section 6. Introduction to the Algebraic Group Purpose of Section: To introduce the concept of a mathematical structure called an algebraic group. To illustrate group concepts, we introduce cyclic and
More information= 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that
Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without
More information9. Quotient Groups Given a group G and a subgroup H, under what circumstances can we find a homomorphism φ: G G ', such that H is the kernel of φ?
9. Quotient Groups Given a group G and a subgroup H, under what circumstances can we find a homomorphism φ: G G ', such that H is the kernel of φ? Clearly a necessary condition is that H is normal in G.
More informationSOLUTIONS FOR PROBLEM SET 6
SOLUTIONS FOR PROBLEM SET 6 A. Suppose that G is a group and that H is a subgroup of G such that [G : H] = 2. Suppose that a,b G, but a H and b H. Prove that ab H. SOLUTION: Since [G : H] = 2, it follows
More informationTheorem Let G be a group and H a normal subgroup of G. Then the operation given by (Q) is well defined.
22. Quotient groups I 22.1. Definition of quotient groups. Let G be a group and H a subgroup of G. Denote by G/H the set of distinct (left) cosets with respect to H. In other words, we list all the cosets
More informationMATH10212 Linear Algebra. Systems of Linear Equations. Definition. An ndimensional vector is a row or a column of n numbers (or letters): a 1.
MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0534405967. Systems of Linear Equations Definition. An ndimensional vector is a row or a column
More information10. Graph Matrices Incidence Matrix
10 Graph Matrices Since a graph is completely determined by specifying either its adjacency structure or its incidence structure, these specifications provide far more efficient ways of representing a
More informationsome algebra prelim solutions
some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no
More informationLECTURE NOTES ON THE MATHEMATICS BEHIND ESCHER S PRINTS: FORM SYMMETRY TO GROUPS
LECTURE NOTES ON THE MATHEMATICS BEHIND ESCHER S PRINTS: FORM SYMMETRY TO GROUPS JULIA PEVTSOVA 1. Introduction to symmetries of the plane 1.1. Rigid motions of the plane. Examples of four plane symmetries
More informationDepartment of Mathematics Exercises G.1: Solutions
Department of Mathematics MT161 Exercises G.1: Solutions 1. We show that a (b c) = (a b) c for all binary strings a, b, c B. So let a = a 1 a 2... a n, b = b 1 b 2... b n and c = c 1 c 2... c n, where
More informationLecture 5 Group actions
Lecture 5 Group actions From last time: 1. A subset H of a group G which is itself a group under the same operation is a subgroup of G. Two ways of identifying if H is a subgroup or not: (a) Check that
More informationThe Topology of Spaces of Triads
The Topology of Spaces of Triads Michael J. Catanzaro August 28, 2009 Abstract Major and minor triads in the diatonic scale can be naturally assembled into a space which turns out to be a torus. We study
More informationGroup Theory: Basic Concepts
Group Theory: Basic Concepts Robert B. Griffiths Version of 9 Feb. 2009 References: EDM = Encyclopedic Dictionary of Mathematics, 2d English edition (MIT, 1987) HNG = T. W. Hungerford: Algebra (SpringerVerlag,
More informationLinear Algebra Concepts
University of Central Florida School of Electrical Engineering Computer Science EGN3420  Engineering Analysis. Fall 2009  dcm Linear Algebra Concepts Vector Spaces To define the concept of a vector
More informationMATH 433 Applied Algebra Lecture 13: Examples of groups.
MATH 433 Applied Algebra Lecture 13: Examples of groups. Abstract groups Definition. A group is a set G, together with a binary operation, that satisfies the following axioms: (G1: closure) for all elements
More informationMT2002 Algebra Edmund F Robertson
MT2002 Algebra 20023 Edmund F Robertson September 3, 2002 2 Contents Contents... 2 About the course... 4 1. Introduction... 4 2. Groups definition and basic properties... 6 3. Modular arithmetic... 8
More informationQuadratic Equations in Finite Fields of Characteristic 2
Quadratic Equations in Finite Fields of Characteristic 2 Klaus Pommerening May 2000 english version February 2012 Quadratic equations over fields of characteristic 2 are solved by the well known quadratic
More informationAssignment 8: Selected Solutions
Section 4.1 Assignment 8: Selected Solutions 1. and 2. Express each permutation as a product of disjoint cycles, and identify their parity. (1) (1,9,2,3)(1,9,6,5)(1,4,8,7)=(1,4,8,7,2,3)(5,9,6), odd; (2)
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationGROUP ALGEBRAS. ANDREI YAFAEV
GROUP ALGEBRAS. ANDREI YAFAEV We will associate a certain algebra to a finite group and prove that it is semisimple. Then we will apply Wedderburn s theory to its study. Definition 0.1. Let G be a finite
More informationNotes on polyhedra and 3dimensional geometry
Notes on polyhedra and 3dimensional geometry Judith Roitman / Jeremy Martin April 23, 2013 1 Polyhedra Threedimensional geometry is a very rich field; this is just a little taste of it. Our main protagonist
More information1 Review of complex numbers
1 Review of complex numbers 1.1 Complex numbers: algebra The set C of complex numbers is formed by adding a square root i of 1 to the set of real numbers: i = 1. Every complex number can be written uniquely
More informationIdeal Class Group and Units
Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals
More informationGroups and Symmetry. Andrew Baker
Groups and Symmetry Andrew Baker [24/12/2005] Department of Mathematics, University of Glasgow. Email address: a.baker@maths.gla.ac.uk URL: http://www.maths.gla.ac.uk/ ajb Contents Chapter 1. Isometries
More information(Q, ), (R, ), (C, ), where the star means without 0, (Q +, ), (R +, ), where the plussign means just positive numbers, and (U, ),
2 Examples of Groups 21 Some infinite abelian groups It is easy to see that the following are infinite abelian groups: Z, +), Q, +), R, +), C, +), where R is the set of real numbers and C is the set of
More informationGroup Fundamentals. Chapter 1. 1.1 Groups and Subgroups. 1.1.1 Definition
Chapter 1 Group Fundamentals 1.1 Groups and Subgroups 1.1.1 Definition A group is a nonempty set G on which there is defined a binary operation (a, b) ab satisfying the following properties. Closure: If
More informationPythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers
Pythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers Amnon Yekutieli Department of Mathematics Ben Gurion University email: amyekut@math.bgu.ac.il Notes available at http://www.math.bgu.ac.il/~amyekut/lectures
More informationd(f(g(p )), f(g(q))) = d(g(p ), g(q)) = d(p, Q).
10.2. (continued) As we did in Example 5, we may compose any two isometries f and g to obtain new mappings f g: E E, (f g)(p ) = f(g(p )), g f: E E, (g f)(p ) = g(f(p )). These are both isometries, the
More information5.3 The Cross Product in R 3
53 The Cross Product in R 3 Definition 531 Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] Then the vector given by [u 2 v 3 u 3 v 2, u 3 v 1 u 1 v 3, u 1 v 2 u 2 v 1 ] is called the cross product (or
More information(a) The transpose of a lower triangular matrix is upper triangular, and the transpose of an upper triangular matrix is lower triangular.
Theorem.7.: (Properties of Triangular Matrices) (a) The transpose of a lower triangular matrix is upper triangular, and the transpose of an upper triangular matrix is lower triangular. (b) The product
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationLecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties
Lecture 1 (Review of High School Math: Functions and Models) Introduction: Numbers and their properties Addition: (1) (Associative law) If a, b, and c are any numbers, then ( ) ( ) (2) (Existence of an
More informationMath 210A: Algebra, Homework 1
Math 210A: Algebra, Homework 1 Ian Coley October 9, 2013 Problem 1. Let a 1, a 2,..., a n be elements of a group G. Define the product of the a i s by induction: a 1 a 2 a n = (a 1 a 2 a n 1 )a n. (a)
More informationChapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
More informationGalois Theory III. 3.1. Splitting fields.
Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where
More informationLinear Dependence Tests
Linear Dependence Tests The book omits a few key tests for checking the linear dependence of vectors. These short notes discuss these tests, as well as the reasoning behind them. Our first test checks
More informationComplex Numbers. Misha Lavrov. ARML Practice 10/7/2012
Complex Numbers Misha Lavrov ARML Practice 10/7/2012 A short theorem Theorem (Complex numbers are weird) 1 = 1. Proof. The obvious identity 1 = 1 can be rewritten as 1 1 = 1 1. Distributing the square
More informationDiagonalisation. Chapter 3. Introduction. Eigenvalues and eigenvectors. Reading. Definitions
Chapter 3 Diagonalisation Eigenvalues and eigenvectors, diagonalisation of a matrix, orthogonal diagonalisation fo symmetric matrices Reading As in the previous chapter, there is no specific essential
More informationWhat are the finite 2dimensional symmetrical objects?
Chapter 2 Groups 2.1 Symmetries Warning! This section attempts to motivate the topic of this chapter, Groups. As a consequence, you will find it vague and you may find it confusing. If you find it too
More information6. ISOMETRIES isometry central isometry translation Theorem 1: Proof:
6. ISOMETRIES 6.1. Isometries Fundamental to the theory of symmetry are the concepts of distance and angle. So we work within R n, considered as an innerproduct space. This is the usual n dimensional
More informationCONSEQUENCES OF THE SYLOW THEOREMS
CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.
More informationSome Results on 2Lifts of Graphs
Some Results on Lifts of Graphs Carsten Peterson Advised by: Anup Rao August 8, 014 1 Ramanujan Graphs Let G be a dregular graph. Every dregular graph has d as an eigenvalue (with multiplicity equal
More informationRevision of ring theory
CHAPTER 1 Revision of ring theory 1.1. Basic definitions and examples In this chapter we will revise and extend some of the results on rings that you have studied on previous courses. A ring is an algebraic
More informationNOTES ON GROUP THEORY
NOTES ON GROUP THEORY Abstract. These are the notes prepared for the course MTH 751 to be offered to the PhD students at IIT Kanpur. Contents 1. Binary Structure 2 2. Group Structure 5 3. Group Actions
More informationDiagonal, Symmetric and Triangular Matrices
Contents 1 Diagonal, Symmetric Triangular Matrices 2 Diagonal Matrices 2.1 Products, Powers Inverses of Diagonal Matrices 2.1.1 Theorem (Powers of Matrices) 2.2 Multiplying Matrices on the Left Right by
More informationDETERMINANTS. b 2. x 2
DETERMINANTS 1 Systems of two equations in two unknowns A system of two equations in two unknowns has the form a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 22 x 2 = b 2 This can be written more concisely in
More informationQuaternions and isometries
6 Quaternions and isometries 6.1 Isometries of Euclidean space Our first objective is to understand isometries of R 3. Definition 6.1.1 A map f : R 3 R 3 is an isometry if it preserves distances; that
More informationABEL S THEOREM IN PROBLEMS AND SOLUTIONS
TeAM YYePG Digitally signed by TeAM YYePG DN: cn=team YYePG, c=us, o=team YYePG, ou=team YYePG, email=yyepg@msn.com Reason: I attest to the accuracy and integrity of this document Date: 2005.01.23 16:28:19
More informationSymmetry Operations and Elements
Symmetry Operations and Elements The goal for this section of the course is to understand how symmetry arguments can be applied to solve physical problems of chemical interest. To achieve this goal we
More informationAlgebra. Sample Solutions for Test 1
EPFL  Section de Mathématiques Algebra Fall semester 20082009 Sample Solutions for Test 1 Question 1 (english, 30 points) 1) Let n 11 13 17. Find the number of units of the ring Z/nZ. 2) Consider the
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in twodimensional space (1) 2x y = 3 describes a line in twodimensional space The coefficients of x and y in the equation
More informationGeometrical symmetry and the fine structure of regular polyhedra
Geometrical symmetry and the fine structure of regular polyhedra Bill Casselman Department of Mathematics University of B.C. cass@math.ubc.ca We shall be concerned with geometrical figures with a high
More informationABSTRACT ALGEBRA. Romyar Sharifi
ABSTRACT ALGEBRA Romyar Sharifi Contents Introduction 7 Part 1. A First Course 11 Chapter 1. Set theory 13 1.1. Sets and functions 13 1.2. Relations 15 1.3. Binary operations 19 Chapter 2. Group theory
More informationSummary of week 8 (Lectures 22, 23 and 24)
WEEK 8 Summary of week 8 (Lectures 22, 23 and 24) This week we completed our discussion of Chapter 5 of [VST] Recall that if V and W are inner product spaces then a linear map T : V W is called an isometry
More informationCOSETS AND LAGRANGE S THEOREM
COSETS AND LAGRANGE S THEOREM KEITH CONRAD 1. Introduction Pick an integer m 0. For a Z, the congruence class a mod m is the set of integers a + mk as k runs over Z. We can write this set as a + mz. This
More informationSECTION 8.3: THE INVERSE OF A SQUARE MATRIX
(Section 8.3: The Inverse of a Square Matrix) 8.47 SECTION 8.3: THE INVERSE OF A SQUARE MATRIX PART A: (REVIEW) THE INVERSE OF A REAL NUMBER If a is a nonzero real number, then aa 1 = a 1 a = 1. a 1, or
More informationDefinition 12 An alternating bilinear form on a vector space V is a map B : V V F such that
4 Exterior algebra 4.1 Lines and 2vectors The time has come now to develop some new linear algebra in order to handle the space of lines in a projective space P (V ). In the projective plane we have seen
More informationSolutions to TOPICS IN ALGEBRA I.N. HERSTEIN. Part II: Group Theory
Solutions to TOPICS IN ALGEBRA I.N. HERSTEIN Part II: Group Theory No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Version: 1.1 Release: Jan 2013
More informationModern Geometry Homework.
Modern Geometry Homework. 1. Rigid motions of the line. Let R be the real numbers. We define the distance between x, y R by where is the usual absolute value. distance between x and y = x y z = { z, z
More informationChapter 10. Abstract algebra
Chapter 10. Abstract algebra C.O.S. Sorzano Biomedical Engineering December 17, 2013 10. Abstract algebra December 17, 2013 1 / 62 Outline 10 Abstract algebra Sets Relations and functions Partitions and
More information