A solvable group that has a composition series is necessarily finite

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Let G be a solvable group. We prove that if G has a composition series, then G has to be finite. (Note that this is sort of a converse to “A finite group has a composition series.”)

Let G=G_0\geq G_1\geq \dots\geq G_n=1 be a composition series of G, where each factor G_i/G_{i+1} is simple.

Since G_i and G_{i+1} are solvable (every subgroup of a solvable group is solvable), the quotient G_i/G_{i+1} is also solvable.

We can prove that G_i/G_{i+1} is abelian. Since (G_i/G_{i+1})'\trianglelefteq G_i/G_{i+1}, by the fact that the factor is simple, we have (G_i/G_{i+1})'=1 or G_i/G_{i+1}.

If (G_i/G_{i+1})'=G_i/G_{i+1}, then this contradicts the fact that G_i/G_{i+1} is solvable. Thus (G_i/G_{i+1})'=1 and G_i/G_{i+1} is abelian.

Key step: G_i/G_{i+1} is simple and abelian, G_i/G_{i+1}\cong\mathbb{Z}_{p_i} for some prime p_i.

Since |G_{n-1}|=p_{n-1}, so we have that |G_{n-2}|=|G_{n-2}/G_{n-1}||G_{n-1}|=p_{n-2}p_{n-1}. By induction, |G_i|=p_i p_{i+1}\dots p_{n-1}.

|G|=|G_0|=p_0p_1\dots p_{n-1}. Thus G is finite.


Author: mathtuition88


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