# Algebraic Methods (November 11, 2011) by Frédérique Oggier

By Frédérique Oggier

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Let us make a few remarks about the definition of solvable group. 47. 1. Every abelian group is solvable. 2. A group G both simple and solvable is cyclic of prime order. 3. A non-abelian simple group G cannot be solvable. Proof. 1. We know that G is abelian if and only if G′ is trivial. We thus get the normal series G(0) = G ⊲ G(1) = {1}. 2. If G is simple, then its only normal subgroups are {1} and G. Since G′ is characteristic and thus normal, we have either G′ = {1} or G′ = G. The latter cannot possibly happen, since then the derived serie cannot reach {1} which contradicts the fact that G is solvable.

The idea of the proof goes as follows: to use the induction hypothesis, we need to get a composition series of length smaller than n, that is, we need to identify the first composition factors, which we will using the above lemma. Concretely, we first exclude the case when G1 = H1 , then compute a composition series of length n − 2 for H1 ∩ G1 , which will indeed be the second composition factor. We then use the second composition series of G to get another composition series for H1 ∩ G1 whose length depends on m, that we can compare to the known one.

Unicity of the composition factors). ) Namely, let Li , i = 1, 2, . . , n denote the distinct terms in the series H1 ⊲ H1 ∩ G1 ⊲ H2 ∩ G1 ⊲ · · · ⊲ Hn ∩ G1 = {1} so that L1 = H1 and L2 = H1 ∩ G1 . Then we have composition series G = H0 ⊲ H1 ⊲ · · · Hn = {1} and G = L0 ⊲ L1 ⊲ · · · Ln = {1} of length n for G and there exists a permutation β of {0, 1, . . , n − 1} such that Hi /Hi+1 ≃ Lβ(i) /Lβ(i)+1 for each i = 0, 1, . . , n − 1. We are almost done but for the fact that we need an isomorphism between Hi /Hi+1 and Gβ(i) /Gβ(i)+1 instead of having Hi /Hi+1 ≃ Lβ(i) /Lβ(i)+1 .