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In his paper "Large Abelian subgroups of $p-$groups", Alperin stated:

Theorem 1: If $p$ is an odd prime and $k$ is a positive integer, then there exists a group of order $p^{3k+2}$ all of whose abelian subgroups have order at most $p^{k+2}$....

In the paragraph after this theorem, he stated

"Burnsides classic theorem: a group of order $p^n$ has (normal) abelian subgroup of order $p^m$ with $n\leq m(m-1)/2$."

As per this Burnsides result, we can say that "A group of order $p^{3k+2}$ contains an abelian subgroup of order $p^{k+3}$, because the inequality in Burnsides classic theorem holds with $n=3k+2$ and $m=k+3$; so how it is possible to get counterexample as in Theorem 1?

Also, as per Burnsides classic result we can say that

"A group of order $p^k$ ($k>4$) contains an abelian subgroup of index $p$; i.e order $p^{k-1}$",

since the inequality in Burnsides result holds for $n=k$ and $m=k-1$ ($k>4$).

Question 1 Can one explain what is correct, what is wrong?

Question 2 Does all maximal abelian normal subgroups of a (non-abelian) $p-$group have same order?

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    You have to read "Burnside's classic theorem" differently: Given a $p$-group of order $p^n$ there exists a normal abelian subgroup of order $p^m$ with $n\le m(m-1)/2$. You wrongly read it as "Given a $p$-group of order $p^n$ there exists for all $m$ with $n\le m(m-1)/2$ a normal abelian subgroup of order $p^m$.", which is clearly absurd as you could choose $m>n$.2011-06-09

2 Answers 2