Abelian subgroup in a 2 group.

Question

Let $G$ be a non-abelian 2-group of order greater than or equal to 32 and $|Z(G)|=4$. Does the group $G$ has an abelian subgroup $H$, such that $16 \leq |H| \leq |G|/2$?

I would also be interested to know how one begins to begin approaching this problem. — Scaramouche, Jul 31 '13 at 05:13
I think the central elements might be useful, but i'm not sure! — D. N., Jul 31 '13 at 05:17
It just came to my mine when i read some properties of groups of order 16 and 32. The answer to the question is positive for groups of order 32. — D. N., Jul 31 '13 at 05:24
@kevin: You mean the bound on $H$? In the above problem, note that 16=32/2. Also abelian subroup of order 8 always exist, namely a subgroup of $\langle x, Z(G) \rangle$, where $x \in G \setminus Z(G)$. — D. N., Jul 31 '13 at 05:58
I am not quite sure why you have the requirement that $|H|\leq |G|/2$ since if it has one of some order, it also has of all smaller orders. Anyway, we can clearly assume that for all $x\in G$, $x^2\in Z(G)$, so $\Phi(G) \leq Z(G)$ and $G'\leq Z(G)$. — Tobias Kildetoft, Jul 31 '13 at 07:52
I don't know how you "clearly assume that for all $x\in G, x^2 \in Z(G)$"?. Can you explain me why? Observe that in the group (of order 32) $\mathbb{Z}_2 \times D_16$, there is an element $y$ such that $y^2 \notin Z(G)$. — D. N., Jul 31 '13 at 08:19
If $x^2$ is not in $Z(G)$ then $\left< x,Z(G)\right>$ has order at least $16$ and we are done. — Tobias Kildetoft, Jul 31 '13 at 08:21
You are going by method of contradiction. Thank you, now i have understand. But what if we want to show that $|H|=|G|/2$? — D. N., Jul 31 '13 at 08:25
I doubt that having center of order $4$ will in general imply the existence of an abelian subgroup whose order is half that of the group, but I have not checked for examples. — Tobias Kildetoft, Jul 31 '13 at 08:32
I doubted too. With your assumption, i notice that the group $G/Z(G)$ is an elementary abelian 2-group. By the way, thank you so much. — D. N., Jul 31 '13 at 08:44
There does not necessarily exist an $H$ with $|H|=|G|/2$. There are many counterexamples of order 64 with centers of order 4, and no abelian subgroups of order 32. — Jack Schmidt, Jul 31 '13 at 14:02

score 3 · Accepted Answer · answered Jul 31 '13 at 08:45

Let $G$ be a group of order greater than or equal to $128$, with $\vert Z(G)\vert=4$. Without loss of generality we may assume that $\vert G\vert =128$. Set $Z=Z(G)$. If $G/Z$ has an element of order $4$, then it has a cyclic subgroup of order $4$ and this corresponds to an abelian subgroup of $G$ of order $16$. Thus as @TobiasKildetoft pointed out, $G/Z$ is an elementary abelian $2$-group, say

$G/Z=\langle a_{1}Z\rangle\times\langle a_{2}Z\rangle\times\langle a_{3}Z\rangle\times\langle a_{4}Z\rangle\times\langle a_{5}Z\rangle$.

for some $a_{1},\ldots,a_{5}\in G\setminus Z$. If $[a_{i},a_{j}]=1$ for $i\neq j$ then we are done as we may take $H=\langle a_{i},a_{j},Z\rangle$. So assume not. Then there exist $i\neq j\in \{2,3,4,5\}$ such that $[a_{1},a_{i}]=[a_{1},a_{j}]$. Equivalently $[a_{j}a_{i}^{-1},a_{1}]=1$. Then taking $H=\langle a_{1},a_{j}a_{i}^{-1},Z\rangle$ gives you an abelian group of order $16$.

You can easily make that argument work for $|G|=64$. If $[a_1,a_2]$, $[a_1,a_3]$, $[a_1,a_4]$ are all different, then $[a_1,a_4]=[a_1,a_2a_3]$, so $[a_1,a_2a_3a_4]=1$. — Derek Holt, Jul 31 '13 at 14:46
@DerekHolt Thanks for that. I thought there was probably a way of doing so, but just couldn't see it! — David Ward, Jul 31 '13 at 15:15

Abelian subgroup in a 2 group.

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