# 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?

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You have to read "Burnside's classic theorem" differently: Given a $p$-group of order $p^n$ there exists a normal abelian subgroup of order $p^m$ with $n\le m(m-1)/2$. You wrongly read it as "Given a $p$-group of order $p^n$ there exists for all $m$ with $n\le m(m-1)/2$ a normal abelian subgroup of order $p^m$.", which is clearly absurd as you could choose $m>n$. – j.p. Jun 9 '11 at 8:04

## 2 Answers

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 $\sqrt{2n}$.

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@Holt:as you said "A group of order $p^n$ has a normal abelian subgroup of order $p^m$ for some $m$, s.t. $n\leq m(m-1)/2$. If this is the case, a group of order $2^5$ will have a normal abelian subgroup of order $2^m$ for some $m$, $5\leq m(m-1)/2$. Such $m$ must be at least $4$,and this would mean a group of order $2^5$ has an abelian normal subgroup of order $2^4$! But this is not true, there are groups of order $2^5$ whose maximal abelian normal subgroup has order $2^3$. If the theorem of Burnside is not giving any existence of at least one such $m$, what is meaning of this "Theorem"? – user8186 Jun 10 '11 at 7:20
@Rahul: Did you read the second paragraph of Derek Holt's answer? – j.p. Jun 10 '11 at 7:31

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.

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