Suppose $G$ is a finite group of order divisible by $8$, with an element $\tau$ of order 2 whose centralizer $C_G(\tau)$ is elementary abelian of order 4. I suspect $G/[G,G]$ must have even order, but I'm not sure how to prove it.
I thought about using transfer and fusion: $\tau$ cannot be conjugate to any involution in the center of a Sylow 2-subgroup (if so, then conjugate that Sylow back so that $\tau$ itself is in the center of a Sylow 2-subgroup, but then that entire Sylow 2-subgroup is contained in the centralizer, contradicting the hypotheses on orders). Since $C_G(\tau)$ contains the center of every Sylow 2-subgroup containing $\tau$ that means that the center of the Sylow 2-subgroups are cyclic of order 2. However, since I don't know much else about the Sylow 2-subgroup, I wasn't sure how to use the transfer.
The goal is to see what sort of classification of groups I can get which have $C_G(\tau)$ of order 4, similar to the one of order 8 mentioned in another question.
Best Answer
If $P$ is a $2$-group with an involution whose centralizer is $2\times 2$, then $P$ is maximal class. This is fairly easy to prove, it's an exercise in I think Huppert. Thus $G$ has either dihedral or semidihedral Sylow $2$-subgroups.
The fast way is to use Thompson's transfer lemma: if $P\in \mathrm{Syl}_2(G)$, $M$ has index $2$ in $P$, and $G$ has no subgroup of index $2$, then all involutions $t$ are $G$-conjugate into $M$. If $P$ is semidihedral then we are now done, because taking $M$ to be the quaternion maximal subgroup means that your involution cannot be conjugate to the central involution. For dihedral $2$-groups you have to choose the dihedral maximal subgroup that does not contain the centralizer.
Alternatively, go directly: if $P\in\mathrm{Syl}_2(G)$ is dihedral then $G/G'$ has no $2$-part if and only if all involutions are conjugate. Since, as you noted, not all involutions are conjugate, $G$ must have a subgroup of index $2$. The same holds if $P$ is semidihedral, since then the involutions in the eihdral maximal subgroup cannot be conjugate, and this means that you can transfer off a $2$ from the top.
So you are correct, you can use fusion to solve this.