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A Some Results from Group Theory A Some Results from Group Theory A.1 Solvable Groups Definition A.1.1. A composition series for a group G is a sequence of sub- groups G = G0 ⊇ G1 ⊇ ...Gk ={1} with Gi a normal subgroup of Gi−1 for each i. A group G is solvable if it has a composition series with each quotient Gi /Gi−1 abelian. Example A.1.2. (1) Every abelian group G is solvable (as we may choose G0 = G and G1 ={1}). (2) If G is the group of Lemma 4.6.1, G = σ, τ | σ p = 1, τ p−1 = 1, τστ−1 = σ r where r is a primitive root mod p, then G is solvable: G = G0 ⊃ G1 ⊃ G2 ={1} with G1 the cyclic subgroup generated by σ . Then G1 is a normal subgroup of G with G/G1 cyclic of order p − 1andG1 cyclic of order p. (3) If G is the group of Lemma 4.6.3, G = σ, τ | σ 4 = 1, τ 2 = 1, −1 3 τστ = σ , then G is solvable: G = G0 ⊃ G1 ⊃ G2 ={1} with G1 the cyclic subgroup generated by σ . Then G1 is a normal subgroup of G with G/G1 cyclic of order 2 and G1 cyclic of order 4. (4) If G is a p-group, then G is solvable. This is Lemma A.2.4 below. (5) If G = Sn, the symmetric group on {1,...,n}, then G is solvable for n ≤ 4. S1 is trivial. S2 is abelian. S3 has the composition series S3 ⊃ A3 ⊃{1}. S4 has the composition series S4 ⊃ A4 ⊃ V4 ⊃{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. 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: 196 A Some Results from Group Theory (1) Gi−1/Gi is solvable for each i. (2) Gi−1/Gi is abelian for each i. (3) Gi−1/Gi is cyclic for each i. (4) Gi−1/Gi is cyclic of prime order for each i. Proof. If Gi−1/Gi is solvable then we may “refine” the original sequence to G ⊇ ··· ⊇ Gi−1 = H0 ⊇ H1 ⊇···⊇Hm = Gi ⊇···⊇Gk ={1} with H j normal in H j−1 and H j−1/H j 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 Gi−1/Gi is cyclic of prime order it is certainly solvable. 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 of G. Then G/N is a solvable group. (3) Let G be a group and let N be a normal subgroup of G. If N and G/N are solvable, then G is solvable. Proof. Consider G = G0 ⊇ G1 ⊇···⊇Gk ⊆{1}. Then H = H0 ⊇ H1 ⊇ ··· ⊇ Hk ⊇{1} where Hi = H ∩ Gi . It is easy to check that Hi is a normal subgroup of Hi−1, and then by standard theorems of group theory Hi−1/Hi = (H ∩ Gi−1)/H ∩ Gi = (H ∩ Gi−1)/(H ∩ Gi−1) ∩ Gi ∼ = (H ∩ Gi−1)Gi /Gi ⊆ Gi−1/Gi is isomorphic to a subgroup of an abelian group so it is abelian. (2) Let π : G → G/N = Q be the quotient map and consider Q = Q ⊇ ∼ 0 ···⊇ Qk ={1} where Qi = π(Gi ). Then Qi = Gi N/N = Gi /Gi ∩ N,so π(Gi−1)/π(Gi ) = (Gi−1 N/N)/(Gi N/N) ∼ = (Gi−1/Gi−1 ∩ N)/(Gi /Gi ∩ N) is isomorphic to a quotient of the abelian group Gi−1/Gi so it is abelian. (3) Let π : G → G/N = Q. Let N = N0 ⊇···⊇N j ={1} and Q = Q0 ⊇ ··· ⊇ Qk ={1} be as in Definition A.1.1. Then the sequence of subgroups −1 −1 −1 G = π (Q0) ⊇ π (Q1) ⊇···⊇π (Qk ) = N0 ⊇ N1 ⊇···⊇N j ={1} shows that G is solvable. A.1 Solvable Groups 197 So far, we have considered abstract solvable groups. However, 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 K of G 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 p-cycle. Regard Sp as operating on T ={1,...,p}. Write T = T1 ∪···∪Tq ,a decomposition onto orbits of K . In other words, i, j, ∈ Tm for some m if and only if there is some τ ∈ K with τ(i) = j. Now for any i, j ∈ T there is a σ ∈ G with σ(i) = j. This implies that the action of G permutes {T1,...,Tq } so in particular each Tm has the same cardinality. This cardinality then divides p,somustbe1orp, and cannot be 1 as K is nontrivial. Thus we see that K operates transitively on T , so has order divisible by p (as the subgroup of K fixing 1, say, has index p). Lemma 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 = G0 ⊃ G1 ⊃···⊃Gk ={1} as in Definition A.1.1. We prove the lemma by induction on k. If k = 1, G = G0 ⊃ G1 ={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 k − 1. Then G1 is a normal subgroup of G, so applying Lemma A.1.5 with K = G1 we see that G1 has order divisible by p, and then by the inductive hypothesis that G1 contains a unique subgroup N of order p. Now let N be any subgroup of G of order p. Then N and N are −1 both p-Sylow subgroups, so are conjugate. Let N = σ Nσ .AsN ⊆ G1, −1 −1 N = σ Nσ ⊆ σ G1σ = G1 as G1 is normal in G,soN = N,as required. Let a group G operate on a set T . This action is effective if the only element σ or G with σ(t) = t for all t ∈ T is σ = id. Let a group G1 operate on a set T1, and a group G2 operate on a set T2. These operations are permutation isomorphic if there is an isomorphism of groups ψ : G1 → G2 and an isomorphism of sets (i.e., a bijection) : T1 → T2 with (σ(t)) = ψ(σ )((t)) for every σ ∈ G1, t ∈ T1. 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 ∗ Fp such that this operation is permutation isomorphic to 198 A Some Results from Group Theory hn G = | h ∈ H, n ∈ F H 01 p operating on i T = | i ∈ F 1 p by left multiplication. Proof. As we have observed, in this situation G has order divisible by p. (Let t ∈ T be arbitrary and consider {σ ∈ G | σ(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 Sp, 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 1 n that N and T are permutation isomorphic to | n ∈ F operating on 01 p i | i ∈ Fp by left multiplication, as a suitable generator ν of N takes 1 i i + 1 to mod p. For simplicity of notation, we simply identify N with 1 1 this group and T with this set. Since N is the unique subgroup of order p of G, it is certainly a normal σ ∈ σνσ−1 = νh = subgroup ofG. Let G be arbitrary. Then for some h 0. 0 n Let σ = . Then, for any i, 1 1 + n ih = νih n = σνi σ −1 n = σνi 0 = σ i , 1 1 1 1 1 so σ i = hn i 1 01 1 and hence hn ∗ G ⊆ | h ∈ F , n ∈ F . 01 p p ∗ Thus, if H is the subgroup of Fp given by A.2 p-Groups 199 ∗ h 0 H = h ∈ F | ∈ G , p 01 then G = G H as claimed. Corollary A.1.8. Let G be a subgroup of Sp, p a prime, of order divisible by p. The following are equivalent: (1) G is solvable. (2) The order of G divides p(p − 1). (3) The order of G is at most p(p − 1). Proof. (1) implies (2): If G is solvable, then, by Proposition A.1.7, G = G H for some H and then |G| divides p(p − 1). (2) implies (3): Trivial. (3) implies (1): G has a p-Sylow subgroup N, and N is unique, as if G had another p-Sylow subgroup N , then G would have order at least p2.
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