I'm trying to work through a sketch proof attributed to Walter Feit on characterizing $S_5$.
Suppose $G$ is a finite group with exactly two conjugacy classes of involutions, with $u_1$ and $u_2$ being representatives. Suppose $C_1=C(u_1)\simeq \langle u_1\rangle\times S_3$ and $C_2=C(u_2)$ be a dihedral group of order $8$. The eventual result is that $G\simeq S_5$. Also, $C(u)$ denotes the centralizer of $u$ in $G$.
The sketch says "$C_2$ contains three classes of involutions. If $x$ is an involution in $C_2$, $x\neq u_2$ then $x$ is conjugate to $xu_2$. Why does $C_2$ contain three classes of involutions, and why is any involution $x\in C_2$ such that $x\neq u_2$ conjugate to $xu_2$?
My thoughts: From the answer here, I know that $C_2$ is a Sylow $2$-subgroup. Since $u_1$ is contained in a Sylow $2$-subgroup, and all the Sylow $2$-subgroups are conjugate to each other, I can replace $u_1$ by some conjugate, and without loss of generality assume that $u_1\in C_2$. Then $u_2\in C_1$ as well.
Then I know $C_2$ contains at least two classes for $u_1$ and $u_2$. They can't be conjugate in $C_2$ lest they be conjugate in $G$. Thus $C_2$ has at least $2$ classes of involutions, but why are there exactly $3$? My guess is $u_1u_2$ might be the third class, but I can't make progress with this guess. Thanks.