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A well known theorem about abelian subgroups of index $p^2$ in $p$-groups is that

(*) if a $p$-group contains abelian subgroup $A$ of index $p^2$, then it contains an abelian normal subgroup $A_1$ of index $p^2$ (see this).

A non-trivial theorem of Alperin asserts that the above theorem is valid if $p^2$ is replaced by $p^3$. Two natural question arise, which I wouldlike to ask here.

  • How long can $A$ and $A_1$ differ in their structure? (in the proof of (*), the normal subgroup $A_1$ obtained is by only considering that $A$ is abelian and without knowing its structure.

  • The second obvious and natural question is- if $G$ contains abelian subgroup of index $p^4$, does $G$ contain abelian normal subgroup of index $p^4$? (and continue question for higher indices)

The second question may be not fully solved, but What is progress on this question?

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First I note that your assertion about Alperin's theorem is not true for $p=2$. I think the answer to your second question is negative. There is $p$-groups ($p>3$) having abelian subgroups of index $p^{\frac{p+3}{2}}$, and no normal abelian subgroup of that index. In particular, for $p=5$ the analogue of Alperin's theorem for the index $p^4$ is false. Such a counter example can be found in Berkovich's book "Groups of prime power order I" (see §39)

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  • $\begingroup$ I do not understand $\frac{p+3}{2}$. The index should be a power of $p$, right? $\endgroup$ Commented Aug 8, 2013 at 21:42
  • $\begingroup$ What do you think? thanks for the remark. $\endgroup$ Commented Aug 8, 2013 at 22:05

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