Abelian subgroups of p-groups 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?
 A: I think Alperin meant the inequality $n \leq m(m+1)/2$.
In a $p$-group $G$ of order $p^n$, any maximal abelian normal subgroup of order $p^m$ has index at most $p^{m(m-1)/2}$.  To see this, observe that a maximal ablian normal subgroup of $G$ is self centralizing, so the quotient $G/A$ can be embedded in $Aut(A)$.  On the other hand the order of $Aut(A)$ can not exceed $p^{d(m-d)} (p^d-1)...(p^d-p^{d-1})$ ( $d$ denotes the rank of $A$) by a well known result of P. Hall, so the order of a p-sylow in $Aut(A)$ is at most $p^{d(m-d)+d(d-1)/2}$ which does not exceed $p^{m(m-1)/2}$, the result follows.
It follows that $n \leq m(m+1)/2$. 
I note also that the above result can be found in Huppert's brilliant book "Endliche Gruppen I", Satz 7.3. 
A: Your interpretation of Burnside's result is incorrect. Alperin's description of Burnside's result would mean that a group of order $p^n$ has a normal abelian subgroup of order $p^m$ for some $m$ satisfying $n \le m(m-1)/2$, not for all such $m$.
But that is clearly not quite right, because it is obviously wrong for $n=3$ and 4. I looked back at Burnside's paper (which I found online at plms.oxfordjournals.org/content/s2-11/1/225.full.pdf), and what he seems to prove is the slightly weaker result that a group of order $p^n$ with centre of order $p^c$ contains a normal abelian subgroup of order $p^m$ for some $m$ with $n \le m + (m-c)(m+c-1)/2$. Burnside also cites a related result of Miller that there is a normal abelian subgroup of order $p^m$, for any $m$ with $n > m(m-1)/2$.
So I think Alperin was being a bit sloppy here. He was really just pointing out that the previous best results had $m$ aprroximately $\sqrt{2n}$.
