Let $p$ be a fixed prime. A group $G$ is termed $p$-central if every element of order $p$ in $G$ lies in the center.

Having a finite $p$-group $G$ of rank $k$ (the least integer, such that every subgroup of $G$ is $k$-generated), one can prove that $G$ has a normal $p$-central subgroup of index bounded in terms of $k$. This can be proved by taking the centralizer $C$ of a maximal normal elementary abelian subgroup $A$ of $G$; the index of $C$ is clearly bounded in terms of the rank of $A$, and so in terms of $k$, and $C$ is $p$-central by a nice result of Alperin (or, in fact, of Thompson and Feit) which says that every element of order $p$ in $G$ which commutes with $A$ lies in $A$.

On the other hand, one can consider the set of all normal subgroups $N$ of $G$, for which $G/N$ is $p$-central. A nice fact is that the quotient of $G$ by the intersection $L(G)$ of all such subgroups, is still $p$-central; so $G/L(G)$ is the unique largest $p$-central quotient of $G$.

Is it true that that the order of $L(G)$ is bounded in terms of $k$ the rank of $G$?

Thanks in advance

  • 1
    $\begingroup$ I think the result you mention requires $p$ to be odd, doesn't it ( I mean the one of Alperin, Feit-Thompson)? $\endgroup$ – Geoff Robinson May 11 '14 at 9:02
  • $\begingroup$ @Geoff Robinson: Yes you are right. For $p=2$, one say that $G$ is $2$-central if every element of order dividing 4 is central. $A$ can be taken to be of exponent 4, so Alperin's result can be applied. $\endgroup$ – Yassine Guerboussa May 11 '14 at 9:22

Let G be a p-group of maximal class and order $p^n\ne2^3$. Then rank of G is at most p. In that case, $L(G)=\Phi(G)$ has order $p^{n-2}$. As $n$ is unbouded, the answer on the question is not. The rank of $G$ is at most $p$ (see Y. Berkovich, Groups of Prime Power Order, 1, Exercise 9.13). If the rank is $p>2$, then $G$ is isomorphic to a Sylow $p$-subgroup $\Sigma_{p^2}$ of the symmetric group of degree $p^2$. For $p=2$ this is not true.

  • $\begingroup$ $G$ depends on $n$ but $\Sigma_{p^2}$ does not... $\endgroup$ – YCor Oct 25 '16 at 17:01
  • $\begingroup$ (However although the formulation is confusing, I guess that that the argument is correct.) $\endgroup$ – YCor Oct 25 '16 at 18:31

Let $H$ be a $p$-group of rank $k$ and define the wreath product $G=H\wr (\mathbf{Z}/p\mathbf{Z})=H^p\rtimes \mathbf{Z}/p\mathbf{Z}$. Then $G=G_H$ has rank at most $pk+1$. The largest quotient of $G$ in which the generator of $\mathbf{Z}/p\mathbf{Z}$ is forced to be central is $H_{\mathrm{ab}}\times\mathbf{Z}/p\mathbf{Z}$. So $L(G)$ has order $\ge |H|^{p-1}$. So taking $H$ of bounded rank and unbounded cardinal is enough, for instance $H$ cyclic of large order.

For instance the group $G_n=\mathbf{Z}/p^n\mathbf{Z}\wr \mathbf{Z}/p\mathbf{Z}$ has order $p^{np+1}$, (bounded) rank $\le p+1$ (actually $p$), and has $L(G_n)$ of (unbounded) order $p^{n(p-1)}$ (and index $p^{n+1}$).

(For $p=2$ it's enough to assume in the definition of $L$ that elements of order 2 are central in $G/L(G)$; if we assume it for elements of order 4 this even makes $L$ larger.)

Also note that choosing $H$ of maximal class, we even get $L(G_H)$ of bounded index $p^3$.


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