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?