Dr. Marques Sophie Algebra Spring Semester 207 Office 59 marques@cims.nyu.edu Problem Set #7 In the following, denotes a isomorphism of groups. Exercise 0 : Let G be a finite abelian group with G n, we will see next week that x n e. Let k ą 0 be an integer such that gcdpk, nq. Prove that every g P G can be written in the form g x k for some x P G. gcdpk, nq ñ Dr, s P Z such that rk ` sn. Now, by Lagrrange s theorem. g n e for all g P G. But, g g g rk`sn pg r q k pg n q s pg r q k.e S pg r q k Take x g r to get x k g. Exercise : In GLpn, Cq and SLpn, Cq define the subgroups of scalar matrices CˆI tλi : λ 0 in Cu Ω n I tλi : λ P Ω n u where Ω n are the complex n th roots of unity. (a) Prove that CˆI and Ω n I are normal in GLpn, Cq and SLpn, Cq respectively. (b) Prove that GLpn, Cq{CˆI SLpn, Cq{Ω n I Hint : Use the Second Isomorphism Theorem. If N CˆI show that l N SLpn, Cq GLpn, Cq (a) If g P λi, (λ 0) then g commutes with every A P GL n pcq, so ApλIqA λi P CˆI for all A P G N pcq and CˆI normal in GL n pcq. Likewise if λ P Ω n, λi now belongs to SL n pcq since detpλiq λ n I I I, and again we have Bpλ IqB λi, @B P SL n pcq ñ Ω n I is normal in SL N pcq. (b) First, note that any A P GL n pcq is λb with detpbq, for a suitably chosen a 0 in C. If n detpaq, it has n th roots λ P C (λ n µ) and then B {λa
has detpbq p{λq n detpaq {µ µ. Thus if N C I, we have N is a normal group of GL n pcq and GL n pcq CˆI SL n pcq. Now apply 2nd Isomorphism theorem taking A SL n pcq, N CˆI. Then A X N SL n pcq X CˆI tλi : λ 0 in C and detpλiq λ n is u That means λ is an n th root of unity, so A X N Ω n I and GL n pcq AN{N» A{pA X Nq SL n pcq{ω n I Exercise 2 : If H is a subgroup of finite index in a group G, prove that there are only finitely many distinct conjugate subgroups aha for a P G. Given a P G, and h P H, the element x ak conjugates H to the conjugates H to the subgroup pahqhpahq ahhh a, since pxyq y x. But hhh H, for all h P H, so pahqhpahq aha for all h P H. The group G is a union of n disjoint cosets a H H, a 2 H,..., a n H, pn G{H q since H has finite index. All x P a n H give the same conjugate xhx, so there are at most n distinct conjugates, H ehe, a Ha 2,..., anha 2 n. Exercise 3 : Let G prˆ, q be the multiplicative group of nonzero real numbers, and let N be the subgroup consisting of the numbers. Let G p0, `8q equipped with multiplication as its group operation. Prove that N is normal in G and that G{N G pr, `q. (a) G is abelian so all subgroups are normal ; to see G{N» G via first Isomorphism theorem. Let φ : G Ñ G be the squaring map f pxq x 2. This is a homomorphism since φpxyq pxyq 2 x 2 y 2 φpxqφpyq. It is surjective since every x ą 0 is φp? xq. Kerpφ t u. By F.I.T, G{N» G. (b) To see G pp0, `8q, q» pr, `q. Taking φpxq lnpxq. This is a bijection and lnpxyq lnpxq ` lnpyq so ln : G Ñ pr, `q is a group». Exercise 4 : If H is a subgroup of G, its normalizer is N G phq tg : ghg Hu. Prove that (a) N G phq is a subgroup. (b) H is a normal subgroup in N G phq. (c) If H Ď K Ď G are subgroups such that H is a normal subgroup in K, prove that K is contained in the normalizer N G phq. (d) A subgroup H is normal in G ô N G phq G. Note : Part (c) shows that N G phq is the largest subgroup of G in which H is normal. 2
(a)trivial. If g, g 2 P N G phq then g g 2 H pg g 2 q g pg 2 Hg 2 qg g Hg H so g g 2 P N G phq. Obviously, g e is in N G phq. Finally, g P NpHq ñ ghg H, ñ H g Hg g Hpg q, so g P N G phq. Finally, g P NpHq ñ ghg H ñ H g Hg g Hpg q so g P NpHq. Done. (b) H is normal in N G phq. Really trivial : g P NpHq ñ ghg H, and clearly H Ď N G phq. (c) Suppose H Ď K are subgroups of G and that H is a normal subgroup K ( khk H, @k P K). Prove that K Ď N G phq. Totally obvious from definition of N G phq. (d) pñq H normal subgroup of G ñ ghg H, @g P G ñ G N G phq. pðq N G phq G ñ ghg H, @g ñ H is normal subgroup of G. Exercise 5 : If x, y P G, products of the form rx, ys xyx y are called commutators and the subgroup they generate rg, Gs x xyx y : x, y P G y is the commutator subgroup of G. Prove that (a) The subgroup rg, Gs is normal in G. (b) The quotient G{rG, Gs is abelian. Hint : In (a) recall that a subgroup H is normal if α g phq ghg Ď H for all g P G. What do conjugations α g do to the generators rx, ys of the commutator subgroup? (a) If x P G, α x pgq xgx takes commutators to commutators : α x pra, bsq α paba b x q xpaba b qx pxax q pxbx q pxpa qx q pxpb qx q α x paqα x pbqα paq x α pbq x rα x paq, α x pbqs rα x pg q pα x pgqq, @g]. Thus each operator α x maps generators of rg, Gs to generators : if S ( the set of all commutators rx, ys, x, y P G) then α x psq Ď S. We must show this ñ α x pă S ąq Ďă S ą, and that will prove normality of ă S ą rg, Gs. In an earlier problem set we showed that the generated subgroup ă S ą for any set S Ď G consists of all words of finite length w a... a r with r ď 8 and a i P S or a i P S. But for any such word, α x pwq α x pa q... α x pa r q is just another word of the same type because if a i s P S, we have α x psq P S, and if a i s for s P S then α x pa i q α x ps q pα x psqq P S. Thus, for @x P G, 3
α x maps words to words, and hence maps ă S ą to ă S ą. Applying this to S pall commutatorsq, we see rg, Gs is normal in G. (b) As for abelian property of the quotient group, let π : G Ñ G{rG, Gs Ḡ be the quotient homomorphism. Then πpaba b q ē, by definition of rg, Gs. But the e πpaqπpbqπpaq πpbq, which implies πpbqπpaq πpaqπpbq. (Elements in rangepπq commute. Since π is surjective, all elements in barg commute. Exercise 7 : Let G be the group of all real 2 ˆ 2 matrices of the form a b such that ad 0. 0 d Show that the commutator subgroup rg, Gs defined in Exercise 3.3.28 is precisely the subset of matrices in G with s on the diagonal and an arbitrary entry in the upper right corner. We do a brute force calculation of a typical ˆ commutator ˆ ABA B, remembering that these a b a b are the generators of rg, Gs. If A, B 0 d 0 d P G. Then ad, a d 0 and d b A {padq, 0 a B {pa d d b q 0 a. All diagonal entries are nonzero. Then by direct matrix calculation b ABA B {d pa bq{pdd q ` pab q{pdd q ` b{d ˆ 0 {pdd qp b d a b ` ab ` bd q ˆ 0 b pa d q{pdd q ` bpd a q{pdd q 0 Take d, d 0 and a such that pd a q{pdd q ; then the set a d (b can be arbitrary in R). ˆ We see that the set S taba B : A, B P Gu contains all elements of the form b C, b P R. 0 b Now rg, Gs ă S ą. But note that S t : b P Ru is already a group 0 b b b under the matrix multiplication( pdetp 0, and 0 0 0 b ` b ). Since ă S ą the smallest subgroup in G that contains the set of generators S, we must have ă S ą S rg, Gs t : b P Ru 0 b 0 Exercise 8 : Consider the group pz{2z, `q. 4
(a) Identify the set of units U 2. (b) What is the order of the multiplicative group pu 2, q? Is this abelian group cyclic? Hint : What is the maximal order of any element g P U 2? (a) In Z{2Z the multiplication units are U 2 trs, r5s, r7s, rsu. (b) U 2 4 ; elements can be have orders opxq, 2, 4 by Lagrange. Now : oprsq ; opr5sq 2; since rs, r5s, r25s rs are pairs as x k opr7sq 2 since rs, r7s, r7s 2 r49s rs oprsq 2 since rs r s and p q 2 rs This group is not cyclic since no x has order opxq 4. Exercise 9 : Let G be any group and let IntpGq be the set of conjugation operations α g pxq gxg on G. Prove that (a) Each map α g is a homomorphism from G Ñ G. (b) Each map α g is a bijection, hence an automorphism in AutpGq. (c) α e id G, the identity map on G. l. If α g pxq gxg then α e pxq exe x, so α e Id G. Also, So, α g g 2 α g α g2. α g g 2 pxq g g 2 xpg g 2 q g pg 2 xg 2 qg α g pα g2 pxqq, @x α g α g pxq α g gpxq α e pxq x which implies α g pα g q. Additional : Show that each α g is an isomorphism G Ñ G (so α g P AutpGqq. Since α g is invertible, it is a bijection, one need only show α g is a homomorphism : α g pxyq gxyg gxg gxg α g pxq α g pyq Exercise 0 : Show that the group IntpGq of inner automorphisms is a normal subgroup in AutpGq. Note : The quotient AutpGq{IntpGq is regarded as the group of outer automorphisms OutpGq. 5
Let α be an arbitrary automorphism and α g P IntpGq. If x P G. Then α α g α pxq αpgα pxqg q αpgqαpα pxqqαpg q αpgq x αpg q αpgq x αpgq α αpgq pxq Therefore α α g α is automorphism. Thus α IntpGq α Ď IntpGq, and IntpGq is a normal sugroup of AutpGq. Exercise : The permutation group G S 3 on three objects has 6 3! elements S 3 te, p2q, p23q, p3q, p23q, p32qu Prove by direct calculation the center of S 3 is trivial (Note : you have proven that G IntpGq). G S 3 ; Show that G» IntpGq. This happens if and only if ZpGq peq. So our problem is to compute ZpS 3 q and show it is trivial. For this permutation group we can list all its elements and compute the 6 ˆ 6 multiplication table shown above. We have S 3 te, p, 2q, p, 3q, p2, 3q, p, 2, 3q, p, 3, 2qu. We omit these routine calculations (they may be simplified by noting that if x p, 2q and y p, 2, 3q. Then p, 3, 2q y and opyq 3 because p, 2, 3qp, 3, 2q p, 3, 2qp, 2, 3q e p, 2, 3q 2 p, 3, 2q p, 2, 3q 3 e Obviously x 2 e, since p, 2qp, 2q e (all the 2-cycle have order 2). Finally, xyx y, by direct calculation. That means S 3 ă x, y ą is isomorphic to the dihedral group D 3, which has trivial center because pn 3 is odd) ) Even if you don t adopt these tricks it is still simple (but tedious to compute the multiplication table ZpGq can be read out of this table as shown above the table. Inspection shows that g e is the only element in S 3. 6
Exercise 2 : For any group G prove that the commutator subgroup rg, Gs ă xyx y x, y P G ą is a characteristic subgroup that is for any σ P AutpGq, we have σprg, Gsq rg, Gs. Hint : What does an automorphism do to the generators of rg, Gs? Note : This example shows that if G is abelian its automorphism gorup may nevertheless be noncommuative (while IntpGq is trivial). If c rx, ys is any commutator in S then αpxq αpxyx y q αpxqαpyqαpxq αpyq rαpxq, αpyqs is just another commutator in S is again in S is again in S because rxys pxyx y q yxy x ry, xs is a commutator in S. Thus S S S Y S. Now rg, Gs ă S ą means : a typical element in rg, Gs is a word g c c 2... c r with r ă 8 and c i P S. Then αpgq αpc q... αpc r q is just another word in rg, Gs so αprg, Gsq Ď rg, Gs. Likewise, taking α in place of α, α rg, Gs Ď rg, Gs which yields the reverse inclusion rg, Gs Ď αrg, Gs Ď rg, Gs. So αrg, Gs rg, Gs as claimed. Exercise 3 : If G is a group, Z is its center, and the quotient group G{Z is cyclic, prove that G must be abelian. Let ā πpaq P G{Z (a P G) be a cyclic generator of G{Z, where π : G Ñ G{Z is the quotient homomorphism. Let A ă a ą in G. The product set A Z is a subgroup in G because z, z P Z Ñ pa z q pazq pa aq pz zq P AZ. Furthermore : πpazq πpaq πpzq and πpzq ē (identity in G{Z) we get πpazq πpaq πta k : k P Zu tpāq k : k P Zu ă ā ą G{Z Thus if g P G, Dx P AZ such that πpxq πpgq, which implies gz xz, and in particular, Dz 0 P Z such that g g e xz 0 P pazq z 0 AZ. Hence, G Ď AZ, so G AZ. If xy P G, we can find a i P A, z i P Z such that x a z, y a 2 z 2. But A ă a ą is 7
obviously abelian (as is any cyclic subgroup) and the z i P ZpGq commute with everybody, so we get G is abelian. xy a z a 2 z 2 a a 2 z z 2 a 2 a z 2 z pa 2 z 2 q pa z q y x 8