Is the largest normal abelian subgroup of a finite 2-group $P$ of order at least the square root of the order of $P$?

Question

Let $G$ be a group of order $2^n$. Does $G$ have a normal abelian subgroup of order at least $2^{n/2}$?

(This is true, via computations in GAP, for $n \le 8$. The question is similar to one posed here: https://math.stackexchange.com/questions/44275/abelian-subgroups-of-p-groups/44283#44283 However, that question, and answer, involves groups of order $p^n$, for odd primes $p$, and I need $p$ to be even!)

About the title: many finite 2-groups don't have a unique largest normal abelian subgroup. — YCor, Commented Aug 7, 2022 at 22:31
I don't know the answer, but George Glauberman wrote many papers about abelian subgroups of finite $p$-groups including about normal ones. — Yiftach Barnea, Commented Aug 8, 2022 at 2:00

Jeremy Rickard · Accepted Answer · 2022-08-08 07:32:22Z

14

In

Alperin, J. L., Large abelian subgroups of p-groups, Trans. Am. Math. Soc. 117, 10-20 (1965). ZBL0132.27204,

the second part of Theorem 1 gives a group of order $2^{50}$ with no abelian subgroups of order greater than $2^{24}$.

answered Aug 8, 2022 at 7:32

Jeremy Rickard

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$\begingroup$ Awesome -- thanks -- that is exactly what I needed! And obviously GAP was not going to get me there! :-) $\endgroup$
– Ken W. Smith
Commented Aug 8, 2022 at 12:23
1

$\begingroup$ @KenW.Smith I have no idea whether that's the minimal example. And of course there may be much smaller examples with no large normal abelian subgroups. $\endgroup$
– Jeremy Rickard
Commented Aug 8, 2022 at 13:57

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