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 $16 \leq |H| \leq |G|/2$?

Answer 1

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 $16$.