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A Chernikov group is a group $G$ that has a normal subgroup $N$ such that $G\over N$ is finite and $N$ is a direct product of finitely many quasicyclic groups.
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$(*)$ A periodic group of automorphisms of a Chernikov group is itself a Cernikov group.

What I have to prove is the following:
- $(*)$ implies that a periodic group of automorphisms of a Chernikov-by-cyclic group is also a Chernikov group.

Let $G$ be a Chernikov-by-cyclic group such that $AutG$ is periodic. Let $C\leq G$ the Chernikov subgroup by hypothesis such that $G\over C$ is cyclic. I have to show that $AutG$ is a Chernikov group.
I work out some infos:
- $AutC$ is a Chernikov group.
- $Aut{G\over C}$ is finite.
- $OutC$ is finite ("Finiteness Condition", vol. 1, Robinson)
- $AutC$ is finite (D. J. S. ROBINSON, Infinite soluble groups with no outer automorphisms, Rend.Sem. Mat. Univ. Padova, 62 (1980), 281-294.)

Clearly a finite group is a Chernikov group.

I miss how connect all this infos... Any ideas?

share|improve this question
    
Why are assuming that ${\rm Aut}(G)$ is periodic (which is not likely to be true)? You are given a periodic group of automorphisms of $G$, which will just be a subgroup of ${\rm Aut}(G)$ in general. My advice would be to read the proof of $(*)$ carefully and try and adapt it to the more general situation. –  Derek Holt Oct 29 '12 at 10:45
    
You're right. I was hoping that with the entire $AutG$ it would be easier; and than, with the sketch of this proof, maybe, would be easier find a more general result. Unfortunaly the proof of $(*)$ is really difficult. I'll give it another try, anyway. –  W4cc0 Oct 29 '12 at 13:45

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