# Transfer and fusion in a centralizer

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

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.

• Thanks! I was trying to figure out what to do when P was not maximal class, but apparently there is no such case :-) May 18 at 15:34
• 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. May 18 at 15:35
• 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 :-) May 18 at 15:58
• 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. May 18 at 16:23