## 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.