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If $G$ is a (non-abelian) $p$-group, $|G|=p^n$, $n>3$, then it is elementary that $G$ contains a (normal) abelian subgroup of order $p^2$. It is also true that $G$ necessarily contains a normal abelian subgroup of order $p^3$ (Group Theory - W. R. Scott).

1) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains a normal abelian subgroup of order $p^m$?

2) What is the largest possible value of $m$ such that any non-abelian group of order $p^n$ contains an abelian subgroup of order $p^m$?

[Please suggest references.]

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    $\begingroup$ Here are some suggested references: G A Miller, On the number of abelian subgroups.. in Messenger Math 36 (1906/7). SC Dancs, Abelian subgroups of finite $p$-groups in Trans AMS 169 (1972). Miller shows a group of order $p^n$ has a normal abelian subgroup of order $p^m$ where $m(m+1)/2 \geq n$. $\endgroup$
    – M T
    Commented Mar 2, 2011 at 11:35
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    $\begingroup$ @mt, your comment should be an answer (after you correct the inequality). $\endgroup$
    – S. Carnahan ♦
    Commented Mar 2, 2011 at 13:11
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    $\begingroup$ books.google.com/… $\endgroup$
    – Steve D
    Commented Mar 2, 2011 at 16:26
  • $\begingroup$ should be tagged p-groups $\endgroup$
    – Alexander Chervov
    Commented Sep 18, 2012 at 6:17

3 Answers 3

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I've been asked to post the following as an answer, although it does not answer either of your questions (i.e. it does not provide the largest $m$).

Here are some suggested references: G A Miller, On the number of abelian subgroups.. in Messenger Math 36 (1906/7). SC Dancs, Abelian subgroups of finite $p$-groups in Trans AMS 169 (1972). Miller shows a group of order $p^n$ has a normal abelian subgroup of order $p^m$ for some $m$ such that $m (m+1)/2 \geq n$. The inequality is correct. Huppert's Endliche Gruppen book is cited there as an alternative proof of Miller's paper (what I remember of that paper is that it is very hard to read).

Edit: a better reference is Zassenhaus's book `The Theory of Groups', IV.3.4. There you find a simple argument for the lower bound above. It's clear from his proof that the lower bound can be improved if you have control of the number of generators of a maximal normal abelian subgroup, for example in the case that the big group is regular.

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  • $\begingroup$ G. A. Miller has lots of amazing results, and a quite unique writing style. I is always nice to get an excuse to read him :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 18, 2011 at 17:13
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George Glauberman and also Jon Alperin and George Glauberman together have written papers on this topic in recent years. One example is: "A note on abelian subgroups of p-groups." Groups St. Andrews 2005. Vol. 2, 445–447, London Math. Soc. Lecture Note Ser., 340, Cambridge Univ. Press, Cambridge, 2007.

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  • $\begingroup$ Yakov Berkovich also has a series of papers on subgroups of finite $p$-groups. For example: Y. Berkovich, On abelian subgroups of $p$-groups, J. Algebra 199 (1998), 262--280. $\endgroup$
    – Primoz
    Commented Apr 18, 2011 at 14:23
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A. Yu. Olshanskii proved in 1978 that, for any $n$ and any prime $p$, there exists a $p$-group of order at least $ p^{{1\over 8}(n^2+4n-8)} $ having no abelian subgroups of order large than $p^n$. Together with Miller's estimate $p^{{1\over2}n(n+1)}$ (see @m_t's answer), this gives quadratic upper and lower bounds.

Similar questions for commutative subalgebras in (associative and Lie) algebras was studied by M.V.Milentyeva. The estimates are also quadratic in this case.

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