Dr. Marques Sophie Algebra Spring Semester 207 Office 59 marques@cims.nyu.edu Problem Set 9 Exercise 0 : Prove that every group of order G 28 must contain a normal subgroup of order 7. (Why should it contain any subgroups of this order?) Hint : If subgroup H has order 7 and H is not normal there is a conjugate H xhx H. What can be said about H X H and the product set H H? Normal means xhx H, @x P G, so there must exist some g P G such that H ghg H. We have H H 7 since conjugation y ÞÑ gyg α g pyq is ijection G Ñ G. By Lagrange, subgroups of G can only have order, 2, 4, 7, 4, 28. Thus the subgroup HXH Ď H can only have HXH or 7 (the divisors of H 7). But if H XH teu, we have (by Extended Lagrange) H H 7 2 49 ą G 28, which is impossible since H H Ď G. We conclude that H must be a normal subgroup in G. Exercise : The units U 9 trks : gcdpk, nq u of the set Z{9Z form a group under the multiplication p q.. Identify the elements in U 9 and make a list shows (a) The element in the subgroup H x ă x ą, (b) The order opxq. for each x P U 9. 2. Identify the cyclic generators of U 9, if any exist, 3. If n ě 3 prove that U n is always even, 4. By examining U n for small n ě 3 provide a counterexample showing that U n ă a ą does not always implies that U n is generated by p q a. Hint : In 3., if n ě 3, then r s rs in U n.
. U 9 trs, r2s, r4s, r5s, r7s, r8su with r8s rs. Generated subgroups are listed below : x opxq ă x ą 2 6, 2, 4, 8, 6 7, 4 5, 0 4 3, 4, 6 7, 28 5 6, 5, 25 7, 35, 5, 25 2, 0 7 3, 7, 49 4, 28 8 2, 8, 64 cyclic generator. 2. x is a generator if and only if opxq U 9 6, so x r2s and x r5s r 4s are the cyclic generators of U 9. 3. The map J : x ÞÑ x r s x is ijection on U N with J 2 Id (so J J ). If n ě 3 then r s rs in U n and Jpxq x all x P U n (because if p q x x for some x, multiplying by x given r s r s xx x x rs and pmod nq n 2.] If U n were odd then would be a partition of U n into pairs tx, Jpx qu,..., tx r, Jpx r qu, with Jpx i q x i, plus are leftover element x r`. But then Jpx r` q could only x r`, a contradiction because Jpx r` q x r` x r`. 4. Trying n 3, 4, 5 we find that U 5 t, 2, 3, 4u has cyclic subgroups such that U 5 4. rs is a cyclic generator but r s r2s r3s is not. x opxq ă x ą 2 4, 2, 4, 8 3, 6 2 3 3, 3, 9 4 4 2, 4, 6 rs is a cyclic generator but r s r2s r3s is not. Exercise 2 : The commutator subgroup rg, Gs of a group G is the subgroup rg, Gs ă S ą generated by the set of all commutators rx, ys xyx y, pall x, y P Gq. Prove that rg, Gs is a normal subgroup in G ; 2. Prove that the quotient group G{rG, Gs is commutative. Hints : For normality of a subgroup H you must show ghg Ď H for every g P G. What does a conjugation x ÞÑ gxg do to a commutator?. A typical element in S is s xyx y for x, y P G. Then gsg gxyx y g gxg gyg gpx qg gpy qg 2
and since x ÞÑ gxg is an isomorphism, we have gxg pgxg q and likewise for y. Thus gsg gxg gyg pgxg q pgyg q rgxg, gyg s which is just another commutator in S, so gsg S for any g P G. Also, note that S S because the inverse of a generator rx, ys is rx, ys pxyx y q py q px q y x yxy x ry, xs (another commutator in S). Thus SYS S. Now, rg, Gs ă S ą implies a typical element u P rg, Gs is a finite-length word u a... a r with a i P S Y S S and r ă 8. Then gug P S Ďă S ą rg, Gs. Thus, for all g P G, we have proved grg, Gsg g ă S ą g Ďă S ą rg, Gs ; 2. Write x for πpxq P G{rG, Gs if x P G. Then xyx y P kerpπq rg, Gs, if x πpxq, ȳ πpyq P Ḡ G{rG, Gs, we get x πpxq πpx q x, ȳ y and xȳp xq pȳq πpxqπpyqπpx qπpy q πpxyx y q πprx, ysq But π bills all commutators since kerpπq rg, Gs, so xȳp xq pȳq ē (the identity in Ḡ). Or xȳ ȳ x, @ x, ȳ P Ḡ. Ḡ is abelian as claimed. Exercise 3 :. If x, y are element in a group G such that they commute that is xy yx and have finite order that is opxq m and opyq n, prove that the order opxyq of their product is always a divisor of the least common multiple lcmpopxq, opyqq lcmpm, nq. 2. If x pi,..., i m q and y pj,..., j n q are disjoints cycles in the permutation group S N, explain why the order of their product xy is always equal to the least common multiple opxyq lcmpm, nq lcmpopxq, opyqq.. Let x, y in a group G such that they commute that is xy yx and have finite order that is opxq m and opyq n, then m lcmpm, nq and n lcmpm, nq with lcmpm, nq m a n b. Thus pxyq lcmpm,nq x lcmpm,nq y lcmpm,nq px m q a py n q b e a e b e (because G is abelian). Then opxyq divides lcmpm, nq. 2. The cycles have order = (length of cycle). So opxq m and opyq n. Since, they have disjoint supports ti,..., i m u and tj,..., j n u they act on different parts of the space r, Ns. Thus pxyq k x k y k (disjoint cycles in S N commute) and the action of pxyq k on u P r, Ns is : pxyq k puq x$ k y k puq x k py k puqq & u i f u R supppxq Y supppyq y k puq i f u P supppyq psince y k puq is also in supppyq.q % x k puq i f u P supppxq psince this implies y k puq uq 3
Now pxyq k e in S n ô pxyq k puq Idpuq u @u P r, ns ô x k puq u @u P supppxq, y k puq u, @u P supppyq (and x k puq y k puq u for u outside supppxq Y supppyq by defn of support ) ô x k Id and y k Id on all of r, ns (x and y act trivially if and only if of their supports) ô x k e and y k e in S n. In particular, if k opxyq, we have x k y k e, so opxq divides k and opyq divides k. lcmpopxq, opyqq divides k opxyq. But in. we showed that opxyq lcmpm, nq. For any a, b 0 in Z, a b and b a ñ b a so opxyq lcmpopxq, opyqq. Exercise 4 : The group G of matrices A b 0 with b P C and detpaq 0 is a subgroup of the full upper triangular group M of matrices B with a, b, c, d P C and detpbq ab 0. Prove that G is normal in M. Is it normal subgroup in the full linear group GL 2 pcq? 2. Prove that the quotient M{G is abelian. t t. If A then 0 A A and G is a group because 0 t s s ` t P G 0 0 0 Then if B we have t {a b{a pa{dq t BAB 0 0 {a 0 which is again in G. Thus G is normal in M. But G is not normal in GL 2 pcq. For instance, if C, we have 0 t 0 a b b CBC 0 0 a b d b ` d which is not in G for all C P GL 2 pcq. 2. If A, A P M we must show paa q G pagq pa Gq pa Gq pagq pa Aq G. This is true if and only if pa Aq paa q G G (the identity coset in M{G) ô 4
pa Aq paa q A pa q A A p the commutator ra, pa q s) is in G. Let A, A d b, so A {detpaq, pa 0 a q {detpa d b q 0 a. Then A pa q AA d b {pdetpaqdetpa d b qq 0 a 0 a 0 d dd {pada d db q ba aa ab ` bd 0 aa d aa {paa dd q dd 0 aa dd 0 The nature of the terms p q, p q does not matter ; it is evident that the product is in the identity coset I G G (ie. is congruent to I (mod G)). Thus M{G is an abelian group. 5