Prove that a finitely generated soluble periodic group is finite.
Let $G=\langle a_1,....,a_k \rangle$
Also, $G$ is soluble, so the derived series for $G$ terminates:
$1 = G^{n} \leq G^{n-1} \leq ... \leq G^{1} \leq G$
Furthermore, I may assume that subgroups of a finite index in a finitely generated group are finitely generated.
Can somebody offer me some insight on this one? I'm having trouble getting started!!
Can you prove it if $G$ is abelian? It's very hands-on.
For solvable $G$ we have a subnormal series $$G=G_n\trianglerighteq G_{n-1} \trianglerighteq \cdots \trianglerighteq G_1 \trianglerighteq G_0 = 1$$ with all factors $G_{i+1}/G_i$ abelian. We proceed by induction on the solvable index $n$. If $n=1$ then $G$ is already abelian and you're done by the "hands-on" exercise.
Suppose $n\geq 2$. Then $G/G_{n-1}$ is finitely-generated, torsion, and abelian, so it is finite. Thus $G_{n-1}$ has finite index in $G$ and is therefore finitely-generated (see this question). On the other hand, $G_{n-1}$ is solvable of index $n-1$, and clearly also torsion, so it is finite by induction. Thus we have a short exact sequence $$1\rightarrow G_{n-1}\rightarrow G\rightarrow G/G_{n-1} \rightarrow 1$$ where the left and right terms are finite. It follows that $G$ is finite.