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.