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Re: Proof "Any solvable group with finite conjugacy classes are finite"
Posted:
Jan 9, 2011 12:36 PM
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On Jan 8, 7:11 pm, Yura Perov <alfu...@gmail.com> wrote:
> Hello! I am trying to solve one simple enough student problem in Group
> Theory.
>
> I should proof, that a solvable group with finite conjugacy classes
> are finite.
I don't think you mean that! All of the conjugacy classes of an
abelian group contain one element only, so any abelian group, finite
or infinite, has finite conjugacy classes.
I think you mean "Prove that a solvable group with finitely many
conjugacy classes is finite".
Prove it by induction on the derived length n of G. Result is obvious
if n=0, and you have proved it yourself when n=1 and G is abelian.
Let K be the last nontrivial term in the derived series. So K is
abelian, and G/K has derived length less than that of G, and hence by
inductive hypothesis G/K is finite. The conjugacy classes of G that
lie in K consist of the orbits of the conjugation action of G on K,
and since K is contained in the kernel of this action, each such orbit
is finite of length at most |G/K|, and now the result follows.
Derek Holt.
> I started with group G, which derivative G' is {e} (the group with an
> one element). If so, then G is the commutative group, and each element
> correspond to the respective conjugacy class. Thereby, G is finite.
>
> If G' != {e}, G'' = {e}, then G' is commutative. In the quotient group
> G/G' there is finite number of conjugacy classes because there exists
> homomorphism from G to G/G', and under the influence of homomorphisms
> conjugate elements turn into conjugate elements.
>
> If I will proof, that G' is finite, I proof that G is finite in this
> case.
>
> So, I should proof, that for any solvable group G with finite
> conjugacy classes G' is finite. I have tried, but for the present I
> haven't found a solution.
>
> Could you help, please, with ideas?
>
> Thank you in advance.
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