Characterization of $A_5$ by the Centralizer of an Involution

In his thesis, Fowler had shown that $A_5$ is the only finite simple group $G$ affording an involution $u$ such that $C_G(u) \cong C_2 \times C_2$. Is there a proof of that result that relies on elementary methods of group theory and representation theory?

For comparison: If a finite simple group $G$ affords an involution $u$ such that $C_G(u) \cong D_4$ is the dihedral group of order 8, then $|G| \in \{ 168, 360 \}$. This is proven in some standard textbooks such as the book of James and Liebeck. The proof relies on the Sylow theorems, the orthogonality relations for characters, and some standard facts about induced characters. Can the above statement be proven by similar methods?

Here are some thoughts:

Let $C = C_G(u)$ be the Klein four group, and let $P \leq G$ be a 2-Sylow subgroup containing $C$. Then $1 \lneq Z(P) \leq C$. Suppose $u \in Z(P)$. Then we have $P=C$. In particular, $P$ has a subgroup of index two containing a unique involution. By a lemma of Thompson [Isaacs - Character Theory of Finite Groups, Lemma 7.11], all involutions of $G$ must be conjugated, whence $P = C_G(x)$ for all $x \in P \setminus \{1\}$. So we are in the situation mentioned by Orat in the comments, and we know how to proceed.

So it remains to consider the case, where $u \notin Z(P)$. Then $Z(P) = \langle z \rangle$ is of order two, and $C = \langle u,z \rangle$. There is a subgroup $D$ between $C$ and $N_P(C)$ isomorphic to the dihedral group of order 8. Of course this situation has to be impossible since it does not match with the $A_5$. Any ideas how to derive a contradiction from here?

I don't know if it helps, but by Brauer-Fowler, $G$ can be embedded into the alternating group $A_{15}$. In particular, the order of $P$ is bounded by $2^{10}$.

I think the remaining case can be handled by exactly the same methods as in the book of Collins:

We have $\sum_i \chi_i(u)^2 = 4$, where $\chi_i$ runs over the irreducible characters of $G$, $\chi_1$ being the trivial character. Because of that equation, there are exactly four of those characters not vanishing on $u$, say $\chi_1, \dots, \chi_4$. Due to orthogonality relations, we may assume $\chi_2(u) = 1$, $\chi_3(u) = -1$ and $\chi_4(u) = \pm 1$. Now, in contrast to the case handled in the book, we have more than one conjugacy class of involutions in $G$. More precisely, $u$ and $z$ cannot be conjugated, since they have centralizers of different orders. Moreover, we also use (as in the book) that the product of two involutions is an involution if and only if they commute. So any two involutions $x,y \in G$ with $xy = u$ must be in $C_G(u)$. It follows that $u$ is not the product of any two elements of its conjugacy class $u^G$. The class algebra constants formula (applied to the conjugacy class $u^G$) yields $$\frac{|G|}{16} \left( 1+ \frac{1}{\chi_2(1)} - \frac{1}{\chi_3(1)} \pm \frac{1}{\chi_4(1)} \right) = 0.$$ Now, since $G$ is perfect, we have $\chi_i(1) \geq 2$ for $i \geq 2$. But then the left hand side is strictly positive. So we arrived at a contradiction here.

I would be happy if anyone can confirm my proof, or point out some mistake.

• For those who will search: I tried to find the thesis online but I can't. I found it in the references of Glauberman (1967) and its title is "Investigations on finite groups of even order." – Orat Nov 28 '17 at 17:50
• @Orat Thank you! I should have mentioned that I also have no access to Fowler's thesis, or to his original proof. – Dune Nov 28 '17 at 17:53
• I also found that similar statement is proved under slightly stronger condition (having $C_2 \times C_2$ as Sylow 2-subgroup) in Collins (1990, p.92). – Orat Nov 28 '17 at 18:26

Theorem: Let $G$ be a finite group of order $2^a n$, where $n$ is odd. If $G$ contains no element with order $2p$ for $p$ an odd prime, and if the $2$-Sylow subgroups $P$ of $G$ are abelian, then one of the following must hold:
(a) $P$ is a normal subgroup of $G$.
(b) $P$ is cyclic and $G$ contains a normal subgroup of index $2^a$.
(c) $G$ is isomorphic to $\operatorname{SL}(2, 2^a)$.