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Let $G$ be a group such that $G'$ abelian and any abelian normal subgroup of $G$ is finite. Show that $G$ is finite.

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  • $\begingroup$ I take it $G'$ is the commutator/derived subgroup of $G$? $\endgroup$ – Elchanan Solomon Jul 29 '13 at 16:33
  • $\begingroup$ "Any abelian normal subgroup" or "any abelian subgroup and any normal subgroup"? $\endgroup$ – Boris Novikov Jul 29 '13 at 16:43
  • $\begingroup$ @BorisNovikov any abelian normal subgroup $\endgroup$ – dimo Jul 29 '13 at 16:49
  • $\begingroup$ So Boris already showed a counterexample, and thus either the claim is false and that's it or else some other condition(s) is(are) lacking... $\endgroup$ – DonAntonio Jul 29 '13 at 17:01
  • $\begingroup$ Perhaps one can show the center is finite index. $\endgroup$ – Jack Schmidt Jul 29 '13 at 19:19
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The centralizer of $G'$ in $G$ has finite index in $G$, so we can assume that it is the whole of $G$: i.e. $G' \le Z(G)$. Let $A=A_1$ be an abelian normal subgroup of $G$ containing $Z(G)$. So $A$ is finite, and hence its centralizer $C_G(A)$ has finite index in $G$. So if $G$ is infinite, then $A$ is properly contained in $C_G(A)$, and then choosing $g \in C_G(A) \setminus A$, we get a larger abelian normal subgroup $A_2 = \langle A,g \rangle$. (It is normal because it contains $G'$.) Hence, if $G$ is infinite, then we can construct an infinite strictly ascending chain $A_1 < A_2 < A_3 < \cdots$ of abelian normal subgroups of $G$, and its union is an infinite abelian normal subgroup, contradiction.

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  • $\begingroup$ I don't understand why you need the stuff with $Z(G)$. To me it seems that your proof works when you just let $A$ be any abelian normal subgroup of $G$ containing $G'$. $\endgroup$ – Mikko Korhonen Jul 29 '13 at 20:07
  • $\begingroup$ Yes you are right! For some reason I had thought the assumption was that $G'$ is finite (rather than $G'$ abelian), so I was imagining that it had to be proved that $G'$ is abelian. $\endgroup$ – Derek Holt Jul 29 '13 at 21:18

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