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The Wielandt-Kegel theorem states that all finite dinilpotent groups G=AB are solvable. This results extends to many infinite groups, e.g., to finitely-generated linear groups. In general however, I think there will be a counterexample. I have searched the literature, but could not find an answer. Is there a (nice, not too pathological ) counterexample ?

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As far as I know, this is still an open question. Why do you think that there is a counterexample? –  Derek Holt Mar 20 '13 at 10:02
    
By Ito's theorem, the Wielandt-Kegel theorem holds for all groups $G=AB$ with $A$ and $B$ abelian, in particular also in the infinite case. This suggests perhaps to start with the case where $G$ is abelian, and $B$ is 2-step nilpotent. –  Dietrich Burde Mar 21 '13 at 16:09

1 Answer 1

up vote 4 down vote accepted

As Derek Holt mentioned, the question is open (if either $A$ or $B$ is not Abelian). Ito's theorem has been generalized several times by N. S. Chernikov and others, for example, see Chernikov, N. S. On groups that are factorizable by two subgroups with Chernikov commutants. Ukraïn. Mat. Zh. 52 (2000), no. 3, 396--402; translation in Ukrainian Math. J. 52 (2000), no. 3, 457–463. A typical result showing the current state of the problem:

A residually finite group $G$ factorized by two subgroups $A$ and $B$ with finite derived subgroups has a normal subgroup $H$ of finite index the second derived subgroup of which is finite and lies in the center $Z( H)$.

If you allow several (but finite number of) factors: $G=A_1A_2...A_n$, then $SL_n(\mathbb{Z})$, $n\ge 3$, is factorizable by cyclic groups (=boundedly generated), and Muranov proved that there are simple 2-generated groups that are boundedly generated. See Carter, D., Keller, G. (1983). Bounded elementary generation of $SL_n(O)$. Am. J. Math. 105(3):673–687, Muranov, Alexey, Diagrams with selection and method for constructing boundedly generated and boundedly simple groups. Comm. Algebra 33 (2005), no. 4, 1217–1258.

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