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.


1 Answer 1


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.

  • $\begingroup$ Thanks! I was trying to figure out what to do when P was not maximal class, but apparently there is no such case :-) $\endgroup$ May 18, 2021 at 15:34
  • 1
    $\begingroup$ Ah, this is a general fact for all primes: if $P$ is a $p$-group and $x\in P$ such that $C_P(x)=p\times p$, then $P$ has maximal class. $\endgroup$ May 18, 2021 at 15:35
  • $\begingroup$ Yes, this is Huppert (Endliche Gruppen I) page 375 Satz 14.23, phrased in terms of size of the conjugacy class. It's also iff! Thanks for bringing this to my attention. I don't remember forgetting it, but that seems like something worth remembering :-) $\endgroup$ May 18, 2021 at 15:58
  • $\begingroup$ Yes, sorry, my fault. I got mixed up. It's proved in Huppert, I think perhaps the only standard place. In Aschbacher or Gorenstein it's an exercise. $\endgroup$ May 18, 2021 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .