hope you all doing fine. I have a question. Is it true that a abelian-by-(finite abelian) group is also (finite abelian)-by-abelian? Thanks.


closed as off-topic by Travis, 6005, Ken, Harish Chandra Rajpoot, Michael Galuza Aug 26 '15 at 5:04

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  • $\begingroup$ What do you mean by that? $\endgroup$ – Robert Israel Aug 25 '15 at 17:27
  • $\begingroup$ You need to say what you mean by an A-by-B group, since there are two opposite conventions for this. $\endgroup$ – Derek Holt Aug 25 '15 at 17:29
  • $\begingroup$ I meant that can we find a finite abelian normal subgroup N such that the quotient G/N is abelian? $\endgroup$ – Yalcin Aug 25 '15 at 17:30
  • $\begingroup$ Oh I am sorry, I am new :) $\endgroup$ – Yalcin Aug 25 '15 at 17:30
  • $\begingroup$ A-by-B means G has a normal subgroup N such that N has the property A, and the quotient G/N has the property B. $\endgroup$ – Yalcin Aug 25 '15 at 17:32

In fact it is not true in either direction.

The infinite dihedral group is abelian-by-finite abelian, but not (finite abelian)-by-abelian.

A central product of infinite many copies of $D_8$ is (finite abelian)-by-abelian, but not abelian-by-(finite abelian).


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