# A solvable group with composition series is finite

I want to prove the following.

A solvable group G with composition series is finite

From the definition of a solvable group I know there exists a normal series:

$\lbrace 1 \rbrace \le G_{n} \leq G_{2} \leq \dots \leq G_{0} = G$

Where each factor is abelian. If this is a composition series I can conclude that each factor is simple and thus simple and abelian which means they must be of the form $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$. From this I can see how we conclude that $G$ must be finite.

So I only need to consider the case when this series is not the composition series and here I'm stuck.

Maybe I'm making some things more complicated. I would appreciate your help/hints.

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You can use iduction on length of composition series. –  mesel Apr 2 '14 at 21:44

## 1 Answer

A subgroup of a solvable group is solvable (Define $H_i=H \cap G_i$, then $H_i$ is a series of $H$ with abelian factors), and a quotient group of a solvable group is solvable (Define $Q_i = G_iN/N$, then $Q_i$ is a series of $Q=G/N$ with abelian factors).

If $K_i$ is a composition series of $G$, then each $K_i/K_{i+1}$ is a quotient of a subgroup of $G$, and so is also solvable. A solvable simple group $F$ is abelian, since $[F,F]$ is a proper normal subgroup of the simple group $F$, and so must be the identity. Hence each composition factor has finite (prime) order. Hence $G$ itself is finite, its order being the product of the orders of its finitely many composition factors.

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(The OP claims to know that a simple abelian group has prime order. It's proven on this site many times of course. A simple abelian group must be cyclic, since a non-identity cyclic subgroup is normal, so the whole group; it must not have any proper subgroups, so it must be prime order.) –  Jack Schmidt Apr 2 '14 at 23:26
Thanks. I guess what I needed were the facts that "a quotient group of a solvable group is solvable" and "A solvable simple group F is abelian". –  HelloProfile Apr 2 '14 at 23:31