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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. The answer 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 '11 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. –  user15767 Jun 14 '11 at 8:38
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This MO question is also relevant: mathoverflow.net/questions/57104 . –  Emil Jeřábek Jun 14 '11 at 10:58
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2 Answers

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.

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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.

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