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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!!

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    $\begingroup$ $G/G^2$ is abelian, and finitely generated, and every element has finite order, hence.... $\endgroup$ – Arturo Magidin May 28 at 1:28
  • $\begingroup$ A direct product of finitely many sylow-p subgroups? $\endgroup$ – Mathematical Mushroom May 28 at 1:57
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    $\begingroup$ 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? $\endgroup$ – Arturo Magidin May 28 at 1:57
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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.

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  • $\begingroup$ Why is it true that $G_{n-1}$ has finite index in $G$ $\rightarrow$ $G_{n-1}$ is finitely generated. $\endgroup$ – Mathematical Mushroom May 28 at 17:10
  • $\begingroup$ 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 $\endgroup$ – Mathematical Mushroom May 28 at 17:11
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    $\begingroup$ 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 ;) $\endgroup$ – Ehsaan May 28 at 17:16
  • $\begingroup$ Actually, I guess I don't need an edit, here is a link $\endgroup$ – Ehsaan May 28 at 17:25

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