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If $G$ is a finite simple group then is it true that an abelian subgroup $H$ of $G$ of maximal order has order $|H| < |G|^{\frac{1}{3}}$? If so, could you please point me to a reference for this, or where a proof is given? Secondly, if $G$ is now a finite nilpotent group of class $2$ and $H$ is again an abelian subgroup of $G$ of maximal order, then is it true that $|H| \leq \sqrt{|G|}$? If so, is there a reference or proof for this?

Thanks in advance for any help.

Sandeep Murthy

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    $\begingroup$ For the second question, just take a direct product of a large abelian group with a small nilpotent group of class 2. $\endgroup$ – Derek Holt Nov 27 '13 at 12:15
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    $\begingroup$ Maybe the following will be usefull: {\bf Theorem.} If $G$ is a group of even order $g > 2$, there exists a proper subgroup $H$ of an order $h > g^{1/3}$. [R.Brauer, Proceedings of the International Congress of Mathematicians, 1954]. $\endgroup$ – Boris Novikov Nov 27 '13 at 15:16
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The answer to both questions is no:

Counterexample to first assertion: $G = {\rm A}_5$, $H = \langle (1,2,3,4,5) \rangle$.

Counterexample to second assertion: $G = \langle (1,2,3,4), (1,3), (5,6) \rangle$, $H = \langle (1,2,3,4), (5,6) \rangle$.

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The result for simple groups is almost true. It's a result of Vdovin:

Theorem: Let $G$ be a finite non-abelian simple group, with $G\not\cong \mathrm{PSL}_2(q)$ for any prime power $q$. Let $A$ be an abelian subgroup of $G$. Then $|G|>|A|^3$.

The result appeared in Algebra and Logic, 38, no. 2 (1999). (Note that any simple group isomorphic to $\mathrm{PSL}_2(q)$ has an abelian subgroup $A$ of order greater than $|G|^{1/3}$: just take $A$ to be a Sylow $p$-subgroup where $p$ is the characteristic of the field.)

Vdovin has another result, which appeared in Algebra and Logic, 39, no. 5 (2000) and which may also be of interest:

Theorem: Let $G$ be a finite non-abelian simple group. Let $N$ be a nilpotent subgroup of $G$. Then $|G|>|N|^2$.

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