Maximal Abelian Subgroups of p-groups

Question

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?

Answer 1

As I explained in my answer on maths.stackexchange, what Alperin wrote is clearly wrong. He has misquoted what Burnside proved, which was 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 $n≤m+(m−c)(m+c−1)/2$. Burnside cites a related result of Miller that there is a normal abelian subgroup of order $p^m$, for any $m$ with $n>m(m−1)/2$. What is that you are still confused about?

The answer to your second question is no. For example a dihedral group of order 16 has a maximal cyclic subgroup of order 8, but it also has subgroups of order 4 isomorphic to $C_2^2$, which are maximal subject to being abelian.

Answer 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 $n \leq m(m+1)/2$.

In particular any p-group of order $p^n$ has an abelian normal subgroup of order $p^m$ such that $n \leq m(m+1)/2$, I think Alperin meant the preceding inequality.

I note also that the above result can be found in Huppert's brilliant book "Endliche Gruppen I", Satz 7.3.