# Prove that a finitely generated soluble periodic group is finite.

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!!

• $G/G^2$ is abelian, and finitely generated, and every element has finite order, hence.... – Arturo Magidin May 28 at 1:28
• A direct product of finitely many sylow-p subgroups? – Mathematical Mushroom May 28 at 1:57
• Well, perhaps, but that is hardly the point... what do you know about finitely generated torsion abelian groups, which is related to what you are trying to prove about soluble/solvable groups? – Arturo Magidin May 28 at 1:57

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.
• Why is it true that $G_{n-1}$ has finite index in $G$ $\rightarrow$ $G_{n-1}$ is finitely generated. – Mathematical Mushroom May 28 at 17:10
• Also, why do the left and right terms in the S.E.S. being finite imply that $G$ is finite? I mean it makes sense but if you could provide the details I'd appreciate it :D – Mathematical Mushroom May 28 at 17:11
• So the claim is that if $H$ is a subgroup of finite index in $G$, and $G$ is finitely-generated, then $H$ is also finitely-generated. I'll edit to include that. For the short exact sequence thing: it's literally a union of cosets, I'll let you think about it ;) – Ehsaan May 28 at 17:16