Solution 9. G. Hiß, M. Künzer. Group theory, WS 07/08. Problem 31

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1 G. Hiß, M. Künzer Problem 31 Group theory, WS 07/08 Solution 9 Note that for h H and g G, we have ξgh) ξg) and ηgh) ηg)h. In particular, ξh) 1 and ηh) h. Moreover, note that ξt) t and ηt) 1 for t T. Suppose given α Z 2 H, A H ). We claim that the map α G H : g, g ) t T α ηgg t)ηg t) 1, ηg t) ) t 1 is in Z 2 G, A). First of all, we note that for g, g G, we get α H G g, 1) t T α ηgt), 1 ) t 1 0 and α H G 1, g ) t T α 1, ηg t) ) t 1 0. Remark. Suppose given h, h, h H. Then αh h 1, h ) αhh 1, h ) + αhh 1, h ) αhh 1, h h 1 )h 0. In fact, this is the factor system condition applied the triple hh 1, h h 1, h ), which proves the remark. Suppose given g, g, g G. We have α G Hg, g )g t g 1 t t T α ηgg t)ηg t) 1, ηg t) ) t 1 g t T α ηgg t)ηg t) 1, ηg t)ηt) 1) ηt)t 1 g t g 1 T α ηgg g t)ηg g t) 1, ηg g t)ηg t) 1) 1 ηg t) t t g 1 T α ηgg g th t )ηg g th t ) 1, ηg g th t )ηg th t ) 1) ηg th t ) th t ) 1 for any tuple of elements h t ) t g 1 T such that h t H. In particular, we may choose h t such that th t T for all t g 1 T. Thus altogether, α G Hg, g )g t T α ηgg g t)ηg g t) 1, ηg g t)ηg t) 1) ηg t)t 1. Therefore α G Hg, g ) α G Hgg, g ) + α G Hg, g g ) α G Hg, g )g t T α ηg g t)ηg t) 1, ηg t) ) α ηgg g t)ηg t) 1, ηg t) ) + α ηgg g t)ηg g t) 1, ηg g t) ) α ηgg g t)ηg g t) 1, ηg g t)ηg t) 1) ) ηg t) t 1 0 by the Remark above. This proves the claim. We claim that α B 2 H, A H ) implies α G H B 2 G, A). Write for some map ρ : H α α ρ : h, h ) ρh ) ρhh ) + ρh)h A such that ρ1) 0. Let ρ G H : G A, g t T ρ ηgt) ) t 1, and note that ρ G H1) t T ρ 1 ) t 1 0.

2 Given g, g G, we have ρ G Hg)g t g 1 t t T ρ ηgt) ) t 1 g t T ρ ηgt)ηt) 1) ηt)t 1 g t g 1 T ρ ηgg t)ηg t) 1) 1 ηg t) t t g 1 T ρ ηgg th t )ηg th t ) 1) ηg th t ) th t ) 1 for any tuple of elements h t ) t g 1 T such that h t H. In particular, we may choose h t such that th t T for all t g 1 T. Thus altogether, ρ G Hg)g t T ρ ηgg t)ηg t) 1) ηg t)t 1 Hence we obtain α ρ G H )g, g ) ρ G Hg ) ρ G Hgg ) + ρ G Hg)g t T ρ ηg t) ) ρ ηgg t) ) + ρ ηgg t)ηg t) 1) ) ηg t) t 1 t T ρ ηg t) ) ρ ηgg t)ηg t) 1 )ηg t) ) + ρ ηgg t)ηg t) 1) ) ηg t) t T α ρ ηgg t)ηg t) 1, ηg t) ) t 1 α ρ ) G Hg, g ), and so α ρ ) G H α ρ G H ) B2 G, A). This proves the claim. The last two claims together with the observation that α α G H is a morphisms of groups from Z 2 H, A H ) to Z 2 G, A) show the welldefinedness of the morphism Cores G H : H 2 H, A H ) H 2 G, A) α + B 2 H, A H ) G α H + B 2 G, A). t 1 Finally, we claim that Cores G H Res G H [G : H] id H 2 G,A), i.e. that α G H G H B 2 G,A) T α for all α Z 2 G, A). Suppose given α Z 2 G, A). Consider the following maps from G to A ρ : g t T α ηgt), t 1) σ : g t T α t 1, g ) and note that ρ1) 0 and σ1) 0. They give rise to the elements α ρ and α σ in B 2 G, A). Calculation 1. We have t T α ηgt), t 1 g ) t g 1 t t T α ηgt)ηt) 1, ηt)t 1 g ) t g 1 T α ηgg t)ηg t) 1, ηg t) t 1) t g 1 T α ηgg th t )ηg th t ) 1, ηg th t ) th t ) 1) for any tuple of elements h t ) t g 1 T such that h t H. In particular, we may choose h t such that th t T for all t g 1 T. Thus altogether, α ηgg t)ηg t) 1, ηg t)t 1), t T α ηgt), t 1 g ) t T

3 finishing this calculation. Calculation 2. We have t T αt 1 g, g ) t g 1 t t T α ηt)t 1 g, g ) t g 1 T α ηg t) t 1, g ) t g 1 T α ηg th t ) th t ) 1, g ) for any tuple of elements h t ) t g 1 T such that h t H. In particular, we may choose h t such that th t T for all t g 1 T. Thus altogether, finishing this calculation. t T αt 1 g, g ) t T α ηgt)t 1, g ), In the following string of equations resp. congruences, we denote maps from G G to A simply by the image of the element g, g ) G G in A. The uncommented equalities mostly follow by the factor system condition on α. α G H G Hg, g ) t T α ηgg t)ηg t) 1, ηg t) ) t 1 t T α ηg t), t 1) t T α ηgg t), t 1) + t T α ηgg t)ηg t) 1, ηg t)t 1) α ρ B 2 G,A) t T α ηgt), t 1) g + t T α ηgg t)ηg t) 1, ηg t)t 1) t T α ηgt), t 1 g ) + t T α ηgt)t 1, g ) t T α t 1, g ) + t T α ηgg t)ηg t) 1, ηg t)t 1) Calc. 1 α σ B 2 G,A) t T α ηgt)t 1, g ) t T α t 1, g ) t T α ηgt)t 1, g ) t T α t 1, gg ) + t T α t 1, g ) g t T α ηgt)t 1, g ) + t T αg, g ) t T αt 1 g, g ) Calc. 2 T αg, g ). This proves the claim. There is a less calculational way to obtain this result; cf. e.g. M. Künzer, Co)homologie von Gruppen, manuscript, RWTH Aachen, The formula for α G H that has been derived in this Problem is stated in the Lemma of the cited paragraph. Problem 32 In this solution, abelian groups are written additively. Replacing B by fb), we may assume B to be a subgroup of A and f to be the inclusion morphism. Let h HomB, Q). We want to show that there exists g HomA, Q) such that g B h. Consider the set M : {C, u) : B C A, u HomC, Q), u B h}. We have B, h) M, so M. Define a partial ordering on M by letting C, u) C, u ) if C C and u C u. Any chain i.e. totally ordered subset) C in M has an upper bound, viz. e.g. C,u) C C, v), where v is defined by the requirement that if C, u) C, then v C u. Note that C,u) C C actually is a subgroup of A. Note that v is a welldefined group morphism. Thus Zorn s Lemma yields a maximal element C 0, u 0 ) in M.

4 Assume that C 0 < A. Let a A C 0. Let C 1 : C 0, a A. Let I : {i Z : ia C 0 }. Write I zz for some z Z, which is possible since I is an ideal in Z and Z is a principal ideal domain. Case z 0. We have C 0, a C 0 a. Hence we may define C u1 1 Q e.g. by u 1 c 0 + ia) : u 0 c 0 ), where c 0 C 0 and i Z. Thus C 0, u 0 ) < C 1, u 1 ), contradicting the maximality of C 0, u 0 ). Case z 0. By divisibility of Q, there exists q Q such that zq u 0 za). Define C 1 u 1 c 0 + ia) : u 0 c 0 ) + iq, where c 0 C 0 and i Z. To show welldefinedness of u 1, let c 0 C 0 and i Z be given such that c 0 + ia c 0 + i a. Then i i )a c 0 c 0 C 0, whence i i jz for some j Z. Hence u 0 c 0 ) + iq) u 0 c 0) + i q) u 0 c 0 c 0) + i i )q u 0 c 0 c 0) + jzq u 0 c 0 c 0) + ju 0 za) u 0 c0 c 0) + jza ) u 0 c0 c 0) + i i )a ) u 0 c0 c 0) + c 0 c 0 ) ) 0 Moreover, u 1 is a morphism of groups. Thus C 0, u 0 ) < C 1, u 1 ), contradicting the maximality of C 0, u 0 ). In both cases, we have reached a contradiction. Thus C 0 A. We may take g : u 0. u1 Q by Note that the case distinction above is not really necessary. The argument for z 0 also works in case z 0, where q can be chosen arbitrarily. We distinguished the cases just to mention that the case z 0 is easy. Lemma 2.55) now follows once more. Suppose given a divisible subgroup B of an abelian group Z. Then id B can be prolonged to a morphism Z ρ B. I.e., calling the embedding B ι Z, we have ιρ id B. Hence the kernel of ρ is a complement to B in Z. Problem 33 1) Suppose given abelian groups U and Ũ and an abelian group C, written additively. We claim that the morphism of groups is an isomorphism. HomU, C) HomŨ, C) HomU Ũ, C) f, f) u, ũ) fu) + fũ) ) Injectivity follows from the fact that f can be recovered from the image of f, f) by precomposition with the inclusion U U Ũ; and similarly f. Let us prove surjectivity. Suppose given a group morphism g : U Ũ C. Let f resp. f be the restriction of g to the first resp. second summand. Then gu, ũ)) gu, 0) + 0, ũ)) gu, 0)) + g0, ũ)) fu) + fũ), showing that g is the image of f, f). This proves the claim. By the Main Theorem on, in this case, finite abelian groups, to prove the assertion of 1), we may assume A to be cyclic, A a, written multiplicatively. Let n be the order of a. Consider the evaluation morphism HomA, C ) C f fa). Since a generates A, it is injective. Its image is given by µ n : {z C : z n 1}, for z n 1 is the only condition on an image z of the generator a that needs to be fulfilled in order to define a group morphism a i z i, where i Z. Now µ n is generated by ζ n : exp2πi/n), which is of order n. Hence altogether HomA, C ) µ n C n A.

5 Note that the assertion does not hold for finitely generated abelian groups in general. In fact, whereas HomZ, C ) C contains elements of any finite order, this does not hold for Z, whence Z HomZ, C ). 2) Recall that C is divisible. By Problem 32, we have a surjective group morphism  ˆB, induced by the inclusion morphism B A. Hence ˆB is isomorphic to a quotient of Â. However,  A and ˆB B by 1). Hence also B is isomorphic to a quotient of A. 3) By 1), it suffices to show that the evaluation map ε : A Â, a λ λa)) is injective. Suppose given a A such that εa) 1, i.e. such that λa) 1 for all λ Â. Assume that a 1. Write A as a direct product of cyclic groups. Let A C B, where C is a cyclic direct factor onto which a projects nontrivially. Define a morphism λ : A C by letting it be injective on C sending a generator of C of order, say, n to an nth primitive root of unity) and trivial on B. Then λa) 1, and we have reached a contradiction. Problem 34 1) We perform an induction on n. If n 0, then G is abelian, and thus G 1 divides p 0 0 1)/2 1. Suppose n 1. Write Z : ZG). Since G/Z is a nontrivial p-group, it has a nontrivial center. Pick x G Z such that xz ZG/Z). Then xz x g Z, and thus [x, g] x 1 x g Z and [g, x] x g x 1 Z for all g G. In particular, h x h xg and x h xg h for all g, h G. Let N : {[x, g] : g G}. As we have just remarked, N Z. We claim that G Z, g [x, g] is a group morphism. Suppose given g, h G. Then This proves the claim. [x, g][x, h] x 1 g 1 xgx 1 h 1 xh x 1 x g x 1 h 1 )h x 1 xg x 1 h 1 )x g h x 1 h 1 x g h x 1 h 1 g 1 xgh [x, gh]. In particular, the image N of this morphism is a subgroup of G. Since N is contained in Z, we even have N G. Moreover, the kernel of this morphism g [x, g] is given by CG x), whence N G / C G x). Since Z C G x) and x C G x) Z, we conclude that N strictly divides G / Z p n, i.e. that N divides p n 1. We have Z/N ZG/N). Moreover, xn Z/N, but xn ZG/N), because [xn, gn] [x, g]n 1 N for all g G. Thus G/N)/ZG/N) strictly divides G/N)/Z/N) G/Z p n, and thus we have G/N)/ZG/N) p m with m < n. By induction, we know that G/N) divides p mm 1)/2 ; in particular, it divides p n 1)n 2)/2. However, G/N) G N/N G /N since N G. So G G /N N, and since G /N divides p n 1)n 2)/2 and N divides p n 1, we can conclude that G divides p nn 1)/2. 2) Let 1 A H P 1 be a central extension such that A MP ) and such that A H ; cf. 2.67), 2.68); in particular, H is a representation group of P, cf. 2.68). Since H/A P and A ZH), the order H/ZH) equals p m for some m n. So we may apply 1) to H to conclude that H divides p mm 1)/2, and thus p nn 1)/2. Now A H shows that A divides H. Altogether, MP ) A divides p nn 1)/2.

6 Christmas Problem Assume the contrary; i.e. assume G to be a finite nonabelian simple group all of whose proper subgroups are abelian. Let M and M be two different maximal proper subgroups of G. Note that M, M G, being a subgroup properly containing M. We claim that M M 1. Suppose given x M M. Then x commutes with all elements of M since M is abelian, and with all elements of M since M is abelian. Hence x commutes with all elements of M, M G. Therefore, x ZG) 1. This proves the claim. For the remainder of the proof, we will only use the claim just established, and that G is a finite nonabelian simple group, to arrive at a contradiction. Note that every element of G is contained in a maximal proper subgroup of G. In fact, any element of G is contained in the cyclic subgroup it generates, which is a proper subgroup of G because G is nonabelian; and this cyclic subgroup in turn is contained in a maximal proper subgroup of G. In other words, G equals the union of all of its maximal proper subgroups. Moreover, 1 is not a maximal proper subgroup of G, for it is contained in a cyclic subgroup generated by some element of G {1}, which is a proper subgroup of G since G is nonabelian. Hence each maximal subgroup of G has order 2. We remark that M N G M), for M N G M) < G, the latter inclusion being strict because M is not normal in G. We claim that any two maximal proper subgroups are conjugate. Assume that M and M are not conjugate. Then ) G g G M g M ) g G g ) {1} g N G M)\G M g {1}) g N G M)\G M g {1})) ) {1} g M\G M g {1}) g M\G M g {1})) 1 + [G : M] M 1) + [G : M] M 1) G G [G : M] [G : M]) G G G /2 G /2) G + 1 by the previous claim, applied to the various maximal subgroups involved, and by the previous remark. As is customary, we have let the symbol g N G M)\G indicate that g runs over a system of representatives of right cosets of N G M) in G; etc. This contradiction proves the claim. By the previous claim, G equals the union of all conjugates of a particular maximal proper subgroup M, so that we obtain G g G M g {1} g N G M)\G M g {1}) {1} g M\G M g {1}) 1 + [G : M] M 1) G + 1 [G : M] G < G. This contradiction solves the Christmas Problem. kuenzer/gt