Review: A homomorphism is a map ' : G Ñ H between groups satisfying 'pg 1 g 2 q 'pg 1 q'pg 2 q for all g 1,g 2 P G. Anisomorphism is homomorphism that is also a bijection. We showed that for any homomorphism ' : G Ñ H, the image and the kernel of the map are subgroups of H and G, respectively: 'pgq imgp'q th P H h 'pgq for some g P Gu H; kerp'q tg P G 'pgq 1 H u G. To check if H Ñ G is a subgroup, (1) check that H contains at least one element; and (2) show that for any x, y P H, you also have that xy 1 P H. (Subgroup criterion) Review: Let A be a non-empty subset of G (not nec. subgroup). The centralizer of A in G is C G paq tg P G gag 1 a for all a P Au. Since gag 1 a Ø ga ag this is the set of elements which commute with all a in A. If A tau, wewritec G ptauq C G paq. Theorem For any non-empty A Ñ G, C A pgq is a subgroup of G. The center of a group G, denoted ZpGq, is the set of elements which commute with everything in G, i.e.zpgq C G pgq. So ZpGq is a subgroup of G of C G paq for all A Ñ G.
More special subgroups Let A Ñ G, andfixg P G. Thendefine ga th P G h ga for some a P Au, Ag th P G h ag for some a P Au, and gag 1 th P G h gag 1 for some a P Au. The normalizer of A in G is the set N G paq tg P G gag 1 Au tg P G ga Agu. Think: The centralizer is the set of g P G that point-wise fixes A by conjugation, whereas the normalizer is the set of g P G that set-wise fixes A by conjugation. Theorem. For any A Ñ G, thenormalizern G paq is a subgroup of G. Moreover, ZpGq C G paq N G paq G.
From last time, conjugation in D 6 looks like y Ñ xyx 1 1 r r 2 s sr sr 2 1 1 r r 2 s sr sr 2 Ò x Ó r 1 r r 2 sr sr 2 s r 2 1 r r 2 sr 2 s sr s 1 r 2 r s sr 2 sr sr 1 r 2 r sr 2 sr s sr 2 1 r 2 r sr s sr 2
Conjugation in D 8 looks like y Ñ xyx 1 1 r r 2 r 3 s sr sr 2 sr 3 1 1 r r 2 r 3 s sr sr 2 sr 3 r 1 r r 2 r 3 sr 2 sr 3 s sr Ò x Ó r 2 1 r r 2 r 3 s sr sr 2 sr 3 r 3 1 r r 2 r 3 sr 2 sr 3 s sr s 1 r 3 r 2 r s sr 3 sr 2 sr sr 1 r 3 r 2 r sr 2 sr s sr 3 sr 2 1 r 3 r 2 r s sr 3 sr 2 sr sr 3 1 r 3 r 2 r sr 2 sr s sr 3
Computing C G paq, ZpGq, and N G paq C G paq tg P G ga ag for all a P Au. ZpGq tg P G gh hg for all h P gu. C G paq tg P G ga Agu. Shortcut: If g commutes with everything in a subset A ta 1,a 2,...,a m u Ñ G, theng commutes with any word a 1 i a 1 1 i a 1 2, where a ij P A. i` So C G paq C G pxayq and N G paq N G pxayq. For example, if S generates G, thenc G psq ZpGq. Conversely, if g 1,g 2,...,g` are all contained in C G paq, ZpGq, or N G paq, thensoeverywordinxg 1,g 2,...,g`y. Example: C D2n pr 2 q, for n 3. First, rpr 2 q pr 2 qr, so xry Ñ C D2n pr 2 q. On the other hand, sr 2 r 2 s r n 2 s, which is r 2 s if and only if n 4. So we have one of the following: 1. If n 4, thens, r P C D8 pr 2 q. So xs, ry C D8 pr 2 q. But xs, ry D 8. So D 8 C D8 pr 2 q. 2. If n 4, thens R C r 2pD 2n q, Moreover, psr m qr 2 r 2 psr m q r n 2 psr m q r 2 psr m q. Since every element of D 2n can be written as r m or sr m for some 0 m n 1, wehave C D2n pr 2 q xry t1,r,r 2,...,r n 1 u.
Example: ZpD 2n q, n 3. Relevant facts are (1) the dihedral group D 2n is generated by ts, ru, and (2) every element of D 2n can be written as r m or sr m for some 0 m n 1. So we need to compute for which m do we have piq pr m qs spr m q and rpr m q pr m qr, and for which m do we have piiq psr m qs spsr m q and rpsr m q psr m qr. Both in (i) are satisfied exactly when r m r m ( r n m ), i.e. when m n m, sothatm n{2. For (ii), the second condition always fails, since rpsr m q sr m 1 and n 3. So ZpD 2n q # t1u t1,r n{2 u if n is odd, if n is even.
More examples (1) If G is abelian, then ZpGq G, C G paq G, N G paq G, for all A Ñ G. (2) We saw in G D 8, ZpGq t1,r 2 u,andfora t1,r,r 2,r 3 u, C G paq A and N G paq G. (3) Claim: in G S n,ifa tpijq 1 1 j nu, then C G paq 1 and N G paq G. Tips for computing 1 : Fact 1: If pa 1 a 2 a m q is a cycle in S n,then pa 1 a 2 a m q 1 p pa 1 q pa 2 q pa m qq. Fact 2: If has cycle decomposition c 1 c 2 c r,then 1 p c 1 1 qp c 1 2 q p c 1 r q. Definition: the cycle type of a permutation is the list (in increasing order) of the cycle lengths in the cycle decomposition of. (More later in Section 4.3) For example, in S 7, the cycle type of p152qp34q is 1, 1, 2, 3.
Mathematical aside: morphisms in general In general in math, a (homo)morphism is just a map from one mathematical object to another of its own kind, which obeys the same rules ( structure-preserving ). (Look up category theory if you ever want to feel like you re doing all math, ever, all at once) Examples: 1. We just defined homomorphisms of groups 2. Linear transformations are morphisms of vector spaces 3. Any map from one set to another is a morphism of sets (no rules!) HompX, Y q t' : X Ñ Y ' is structure-preservingu Endomorphisms are morphisms from something to itself. EndpXq HompX, Xq t' : X Ñ Xu Isomorphisms are bijective morphisms. Automorphisms are bijective endomorphisms. AutpXq t' : X Ø Xu Group actions Goal: Build automorphisms of sets (permutations) by using groups. Fact: AutpAq S A for a set A. Examples we already know: 1. The dihedral group permutes the set of symmetric states a regular n-gon can occupy. 2. The symmetric group S X permutes the objects of X. 3. The invertible matrices permute vectors in a vector space. 4. The symmetric group S n can also move vectors in R n around via permutation matrices. (Recall representations) Basically, if A is the set we re permuting, then the goal is to find homomorphisms from G into S A.
Group actions: new language, same idea. A group action of a group G on a set A is a map from G ˆ A pg, aq Ñ A fiñ g a that satisfies g ph aq pghq a and 1 a a (structure preserving) for all g, h P G, a P A. We say G acts on A, denoted G ü A.
Two dihedral examples: (1) The dihedral group D 8 acts on r4s t1, 2, 3, 4u as follows: Label the vertices of the square. For g P D 8,let g be the permutation given by g piq is the vertex appearing in i s place after applying the symmetry g. 1 2 2 3 3 4 4 1 4 3 1 1 1 4 r p1234q 2 1 r 2 p13qp24q 3 2 r 3 p1432q 2 1 3 2 4 3 1 4 3 4 s p12qp34q 4 1 rs p13q 1 2 r 2 s p14qp23q 2 3 r 3 s p24q (2) Now consider the induced action on size two subsets of r4s. t1, 2u t1, 3u t1, 4u t2, 3u t2, 4u t3, 4u 1 t1, 2u t1, 3u t1, 4u t2, 3u t2, 4u t3, 4u r t2, 3u t2, 4u t1, 2u t3, 4u t1, 3u t1.4u r 2 t3, 4u t1, 3u t2, 3u t1, 4u t2, 4u t1, 2u r 3 t1, 4u t2, 4u t3, 4u t1, 2u t1, 3u t2, 3u s t1, 2u t2, 4u t2, 3u t1, 4u t1, 3u t3, 4u sr t1, 4u t1, 3u t1, 2u t3, 4u t2, 4u t2, 3u sr 2 t3, 4u t2, 4u t1, 4u t2, 3u t1, 3u t1, 2u sr 3 t2, 3u t1, 3u t3, 4u t1, 2u t2, 4u t1, 4u
New vocabulary for old examples: 1. The dihedral group acts on the set of symmetric states a regular n-gon can occupy by rotations and flips. 2. The symmetric group S X acts on X by permutation. 3. The invertible matrices act on vector spaces. 4. The symmetric group S n also acts on R n by permutation matrices. More examples: 1. Rˆ acts on R n by scaling: x pv 1,v 2,...,v n q pxv 1,xv 2,...,xv n q. 2. Any group G acts on itself (let A G) inseveralways: left regular action: g a ga right multiplication: g a ag 1 conjugation: g a gag 1 Note: The way we ve been writing the action (g a) iscalleda left action. Sometimes it can be better to write a g means g is acting from the right. The right regular action is a g ag. (Di erent from the left action of right multiplication!)