A non-abelian group of order $p^n$ ($n\geq 4$) always has normal abelian group of order $p^3$, and this theorem is useful in enumeration/ classification of groups of order $p^4$. So, abelian normal subgroups of $p$ groups are useful in the classification problem.

Alperin, in his paper on "Large Abelian Subgroups of $p$ groups" stated a result of Burnside namely

"*a group of order $p^n$ has normal abelian subgroups of order $p^m$ with $n\leq m(m-1)/2$*".

Question: For (non-abelian) group $G$ of order $2^5$, by result of Burnside, there will be normal abelian subgroups of order $p^m$ with $5\leq m(m-1)/2$, which means $m\geq 4$. So conclusion is $G$ always has normal abelian subgroup of order $2^4$. But if we check the list of groups of order $2^5$, then there are some non-abelian groups where maximaum order of abelian (normal) subgroup is $2^3$.

Can one explain, what is going wrong here? (I am confused with this theorem.)

Does all maximal abelian subgroups of a non-abelian finite $p$ group have same order?

Also, please, suggest some reference for some results on maximal abelian subgroups of $p$ groups?