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May 19, 2013 at 17:36 comment added Yassine Guerboussa I think this example is due to A. Mann (with some modification, so be sure that any mistake in the above example is due to Y. Guerboussa)
May 19, 2013 at 17:28 comment added Yassine Guerboussa To see this, pick two elements $x$ and $y$ of $K$ one in the center of $K$ (not in $Z(G)$) and the other not. In the vector space $F/H$, we can construct an automorphism which permutes $xH$ and $yH$, and which leaves $K/H$ invariant, this can be lifted to an automorphism of $K/U$ (since $G$ is relatively free), and so $Z(K)$ is not invariant under this automorphism, a contradiction. Perhaps, this may help in constructing other examples.
May 19, 2013 at 17:12 comment added Yassine Guerboussa For any $n>2$ consider the relatively free group $G_n = F_n/U$, where $F_n$ is the free group on n-generators and $U = [F,F,F]F^p$. It follows that $Z(G)$ is equal to $H/U$ where $H = [F,F]F^p$ (since $H/U$ is the largest characteristic subgroup of $G_n)$. For any non-abelian subgroup $K$ of $G$ containing $Z(G)$, we have $Z(G)=Z(K)$ and so this hold for $K$ maximal.
May 18, 2013 at 13:36 history asked Yassine Guerboussa CC BY-SA 3.0