15.24 (2002)
SolvedSuppose that a finite $p$-group $G$ has a subgroup of exponent $p$ and order $p^n$. Is it true that if $p$ is sufficiently large relative to $n$, then $G$ contains a normal subgroup of exponent $p$ and order $p^n$?
Progress
J. L. Alperin and G. Glauberman (J. Algebra, 203, no. 2 (1998), 533–566) proved that if a finite $p$-group contains an elementary abelian subgroup of order $p^n$, then it contains a normal elementary abelian subgroup of the same order provided $p > 4n-7$, and the analogue for arbitrary abelian subgroups is proved in (G. Glauberman, J. Algebra, 272 (2004), 128–153).
Yes, it is true if $p > n$. By induction, $G$ contains a subgroup $H$ of exponent $p$ and order $p^n$ which is normal in a maximal subgroup $M$ of $G$. Then $H \leqslant \zeta_{p-1}(M)$. The elements of order $p$ of $\zeta_{p-1}(M)$ constitute a normal subgroup of $G$, which contains $H$. (A. Mann, Letter of 1 October 2002.)
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