# Involution centralizer of perfect group with quaternion Sylow 2-subgroups

Exercise 7.11 of Kurzweil–Stellmacher asks me to:

Prove that the centralizer of an involution is nonsolvable inside any perfect group with generalized quaternion Sylow 2-subgroups of order at least 16.

I would like to oblige, but I'm having trouble getting much of a handle on the centralizer itself, rather than just its action.

The center T=⟨t⟩ of a quaternion group is weakly closed, and so the normalizer X=NG(T)=CG(t) of T in that perfect group G controls G-fusion within any Sylow 2-subgroup P that happens to contain T.

The only fusion possible on P in a perfect group is the "full" fusion where all (both classes of) quaternion subgroups of order 8 are acted on by their full automorphism group (and otherwise no fusion beyond that induced by P itself).

So I know X contains at least two copies of SL(2,3) and X/Z contains at least two copies of S4. Since G is perfect, P = [P,X] ≤ [X,X]. I think the copies of SL(2,3) pairwise intersect in a particular cyclic subgroup of P of order 4.

More precisely: If Q is a quaternion subgroup of P of order 8, then Y=NX(Q) contains SL(2,3) and Y/Z contains S4. [Y,Q] ≥ Q. P is generated by its [T:Q] quaternion subgroups of order 8, so also by its [Y,Q]s. The intersection of any pair of distinct Qs is the order 4 cyclic subgroup of the (unique) maximal subgroup that happens to be cyclic.

So it is a special configuration that has to fit lots of solvable groups inside, but I don't see any particular reason the result could not be solvable. The closest thing to nonsolvable I get is that the derived length of N has to be at least 3.

I get roughly the same results for a quaternion Sylow of order 8, except one only has the single SL(2,3). I have not found a perfect G with solvable X in this case, but of course I can if I only require the same fusion pattern, since SL(2,3) and SL(2,5) have the same 2-fusion.

More: With a little reminder from reading the work of Bender's students, I can now do this using Brauer–Suzuki or Z*, but this does not seem much like a homework solution.

The truth seems to be X⋅O(G)=G, and every non-trivial quotient of a perfect group is perfect, so the homomorphic image G/O(G) of X is perfect. Since G has Sylow 2-subgroups, G/O(G) is non-identity, so X has a non-identity perfect quotient, so X is nonsolvable.

I'm not sure how to show X⋅O(G)=G without using tools significantly beyond chapter 7.

Smaller question: I vaguely recall Suzuki proved that if G is a perfect group with no non-identity normal subgroups of odd order and with quaternion Sylow 2-subgroups, then each coset of CG(t) contains exactly one conjugate of t, where t is an involution. In other words, if the product of two involutions centralizes t, then that product is the identity.

Can someone prove this with ideas from K–S up to chapter 7?

This is hard for me to check since in reality such a group has exactly one involution t and G = CG(t), but that comes later in my solution.

• The smaller question is answered affirmatively by the proof of K–S 8.1.8, which could easily be "up to chapter 7", as it is just fiddling with dihedral groups and commutators. – Jack Schmidt Dec 4 '10 at 20:37

Let $P$ be generated by $x$ of order $2^n > 4$ and $y$ of order $4$, and let $z = x^{2^{n-2}}$ be a power of $x$ of order 4. Then the two classes of $Q_8$ in $P$ are represented by $\langle z,y \rangle$ and $\langle z, xy \rangle$.
From what you say, you know that both of these subgroups are normalized in $X$ by subgroups isomorphic to ${\rm SL}(2,3)$. That means that $z$ is conjugate in $X$ to both $y$ and to $xy$. I don't think that's possible in a solvable group $X$.
Assume that $X$ is solvable, so it has a chief series with elementary abelian factors. So $z \in M \setminus N$ for one of these factors $M/N$. If $z$ is conjugate to both $y$ and $xy$, then we have $y, xy \in M \setminus N$. But then $x \in M$ and so $x^2 \in N$ and hence $z \in N$, contradiction.
ADDED: For your smaller question if, for conjugates $u$ and $v$ of $t$, $x=uv$ is nontrivial and centralizes $t$, then $x$ has odd order and both $t$ and $u$ lie in the normalizer $N$ of $\langle x \rangle$ and hence are conjugate in $N$, but that is impossible because $t$ centralizes and $u$ inverts $x$.