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Some Results from Group Theory

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Some Results from Group Theory Some Results from Group Theory A.l Solvable Groups Definition A.1.1. A composition series for a group G is a sequence of sub­ groups G = Go2Gi 2...GA: = {1} with G/ a normal subgroup of G/_i for each /. A group G is solvable if it has a composition series with each quotient Gi/Gi-i abehan. o Example A.1.2. (1) Every abeUan group G is solvable (as we may choose Go = GandGi = {1}). (2) If G is the group of Lemma 4.6.1, G = (a, r | a^ = 1, r^~^ = 1, rar"^ = <j^} where r is a primitive root mod/?, then G is solvable: G = GODGIDG2 = {1} with Gi the cyclic subgroup generated by a. Then Gi is a normal subgroup of G with G/Gi cyclic of order p — 1 and Gi cyclic of order p. (3) If G is the group of Lemma 4.6.3, G = (a, r | a"^ = 1, r^ = 1, rorr"^ = a^) , then G is solvable: G = Go D Gi D G2 = {1} with Gi the cyclic subgroup generated by a. Then Gi is a normal subgroup of G with G/Gi cyclic of order 2 and Gi cyclic of order 4. (4) If G is a /?-group, then G is solvable. This is Lenmia A.2.4 below. (5) If G = Sn, the synmietric group on {1,..., n}, then G is solvable for n < 4. ^1 is trivial. S2 is abelian. ^3 has the composition series ^3 D A3 D {1}. 54 has the composition series ^4 D A4 D V4 D {1} where V4 = {1, (12)(34), (13)(24), (14)(23)}. (Here An denotes the alternating group.) (6) If G = Sn or G = An, then G is not solvable for n > 5. This is Corollary A.3.6 below. o Lemma A.1.3. A finite group G is solvable if and only if it has a composition series satisfying any one of the following conditions: 170 A Some Results from Group Theory (1) Gi-i/Gi is solvable for each i. (2) Gi-i/Gi is abelianfor each i. (3) Gi-i/Gi is cyclic for each i. (4) Gi-i/Gi is cyclic of prime order for each i. Proof If Gi-i/Gi is solvable then we may "refine" the original sequence to G 2 • • 2 Gi-i = Ho 2 ^1 2 • • • 2 ^m = G/ 2 • • 2 G^ = {1} with Hj normal in Hj-i and Hj-i/Hj abelian; similarly we may refine a sequence with abelian quotients to one with cyclic quotients and one with cyclic quotients to one with quotients cyclic of prime order. On the other hand, if Gt-i/Gt is cyclic of prime order it is certainly solvable. n Lemma A.1.4. (1) Let G be a solvable group and let H be a subgroup of G. Then H is a solvable group, (2) Let G be a solvable group and let N be a normal subgroup ofG. Then G/N is a solvable group, (3) Let G be a group and let N be a normal subgroup ofG.IfN and G/N are solvable, then G is solvable. Proof Consider G = Go 2 Gi 2 • • • 2 G^ c {1}. Then H = HQ ^ Hi ^ • • • 2 ^A: 2 {1} where Hi = H D G/. It is easy to check that Hi is a normal subgroup of Hi-I, and then by standard theorems of group theory Hi_,/Hi = (Hn Gi.i)/H n Gi = (Hn Gi-i)/(H n G,_i) n G, = (//nG,_i)G,/G,CG,_i/G, is isomorphic to a subgroup of an abelian group so it is abelian. (2) Let 7t:G-^ G/N = g be the quotient map and consider Q = QQ :D ... 2 e^ = {1} where Qi = n{Gi), Then Qi = GiN/N = Gi/Gi n N, so 7r(G,_i)/7r(G,) = (Gi.iN/N)/iGiN/N) = (Gi-i/Gi-inN)/(Gi/GinN) is isomorphic to a quotient of the abelian group Gi-i/Gi so it is abelian. (3) Let TT: G -^ G/N = Q, Lei N = No ^ -" ^ Nj = {1} and G = Go 2 • • • 2 2A: = {1} be as in Definition A. 1.1. Then the sequence of subgroups G = 7c-\Qo) 2 7t-\Qi) 2 • • • 2 yr'^Qk) = iVo 2 M 2 • • • 2 iV^ = {1} shows that G is solvable. n A. 1 Solvable Groups 171 So far, we have considered abstract solvable groups. Hov^ever, when we consider solvable subgroups of symmetric groups, we can get very precise descriptions. Lemma A.1.5. Let G be a subgroup of Sp of order divisible by p, p a prime. Then any nontrivial normal subgroup KofG has order divisible by p. Proof Since G has order divisible by p, it must contain an element of order p, which must be a /?-cycle. Regard Sp as operating on T = {1,..., p]. Write T = Ti U • • • U T^, a decomposition onto orbits of K, In other words, /, y, G Tm for some m if and only if there is some x e K with x{i) = j. Now for any i, j e T there is a a e G with or(/) = j. This implies that the action of G permutes {Ti,..., T^} so in particular each T^ has the same cardinality. This cardinality then divides p, so must be 1 or /?, and cannot be 1 as ^ is nontrivial. Thus we see that K operates transitively on 7, so has order divisible by p (as the subgroup of K fixing 1, say, has index p). D Lenvma A.1.6. Let G be a solvable subgroup of Sp of order divisible by p, p a prime. Then G contains a unique subgroup N of order p. Proof Let G = Go D Gi D • • D Gjt = {1} as in Definition A.LL We prove the lenmia by induction on k. \fk = 1, G = Go D Gi = {1} and G is solvable, so G is abelian, and hence has a unique p-Sylow subgroup N, of order p. Suppose the lemma is true for A: — L Then Gi is a normal subgroup of G, so applying Lemma A.L5 with A' = Gi we see that G\ has order divisible by p, and then by the inductive hypothesis that G\ contains a unique subgroup N of order p. Now let A^' be any subgroup of G of order p. Then N' and N are both p-Sylow subgroups, so are conjugate. Let A^' = aNa~^. As N c Gi, A^' = aNcr~^ c aGia~^ = Gi as Gi is normal in G, so N^ = N, as required. D Let a group G operate on a set T. This action is effective if the only ele­ ment a or G with a(t) = ^ for all ^ G T is a = id. Let a group Gi operate on a set Ti, and a group G2 operate on a set T2. These operations are permutation isomorphic if there is an isomorphism of groups xlr: Gi ^ G2 and an isomorphism of sets (i.e., a bijection) *: Ti -> T2 with *(a(0) = ^(or)(vl/(0) for every a eGut e Ti. Proposition A.1.7. Let G be a solvable group operating effectively and tran­ sitively on a set T of cardinality p, p a prime. Then there is a subgroup H of F* such that this operation is permutation isomorphic to 172 A Some Results from Group Theory ^^ = l[oi]' ^e^'^eFp} operating on =![;]'•• €F,) by left multiplication. Proof. As we have observed, in this situation G has order divisible by p. (Let ^ € r be arbitrary and consider [a e G \ a{t) = t]. Since G acts transitively on T, this subgroup of G has index p.) Since G operates effectively on T, G is isomorphic to a subgroup of the symmetric group 5^,, so, by Lemma A. 1.6, G contains a unique subgroup N of order p, which is generated by a p-cycle. Reordering, if necessary, we have that A^ and T are permutation isomorphic to j L . | n e F^ I operating on I I / € Fp I by left multiplication, as a suitable generator v of N takes mod p. For simplicity of notation, we simply identify N with thiMV]s group and T with this set. Since A^ is the unique subgroup of order p of G, it is certainly a normal subgroup of G. Let a € G be arbitrary. Then ava~^ = v^ for some h ^ 0. Let Then, for any /, so ^([;])=[o"][i] and hence heV;, neFpY Thus, if H is the subgroup of F* given by A.2 /7-Groups 173 A.=|*^ni[*„?]^4 then G = GH as claimed. n Corollary A.1.8. Let G be a subgroup of Spy p a prime, of order divisible by p. The following are equivalent: (1) G is solvable. (2) The order ofG divides p(p — 1). (3) The order ofG is at most p(p — I). Proof (1) implies (2): If G is solvable, then, by Proposition A. 1.7, G = GH for some H and then \G\ divides p(p — I). (2) implies (3): Trivial. (3) implies (1): G has a /7-Sylow subgroup A^, and N is unique, as if G had another /?-Sylow subgroup N\ then G would have order at least p^. Then the last part of the proof of Proposition A. 1.7 shows that G = GH for some H, But GH ::> N D {1} with A^ normal in G. N is abeUan and GH/N is isomorphic to H which is abelian, so G is solvable.
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