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
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: $G = \langle (1,2,3,4), (1,3), (5,6) \rangle$, $H = \langle (1,2,3,4), (5,6) \rangle$.
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 $|G|>|N|^2$.
