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What can one say about a finite p-group $G$ in which the center of any maximal subgroup is equal to the center of $G$?. CH. J.Cossey and T.Hawkes, (Sets of p-powers as conjugacy class sizes, Proc. Amer. Math. Soc. 128 (2000), 49-51.) proved that for any finite set $S$ of powers of $p$, including $1$, there exists a p-group (of class 2) whose conjugacy class sizes are exactly the members of $S$. It follows from this that there exists a p-group (of class 2) satisfying the property in the question. I ask particularly if one can find such a p-group with class greater than 2.

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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. – Yassine Guerboussa May 19 '13 at 17:12
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. – Yassine Guerboussa May 19 '13 at 17:28
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) – Yassine Guerboussa May 19 '13 at 17:36

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