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!!

  • 2
    $\begingroup$ $G/G^2$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.