## Maximal Abelian Subgroups of p-groups

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?

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A very similar question can be found at [math.stackexchange.com](math.stackexchange.com/questions/44275/…). The [answer](math.stackexchange.com/questions/44275/…) by Derek Holt to the first question is quite good. It might be better, if you restrict yourself to asking only one question per question (I see 3 questions here). – Someone Jun 14 2011 at 8:34
@Someone: I went through some papers of Alperin and Burnside, but still I am not satisfied. I didn't get enough material. If someone gives some direction for these questions, then its fine. – unknown (google) Jun 14 2011 at 8:38
This MO question is also relevant: mathoverflow.net/questions/57104 . – Emil Jeřábek Jun 14 2011 at 10:58

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.