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 $G$.
I'm trying to deduce that $|S_2|=0$ or $4$, where $S_i$ is the set of pairs $(x,y)$ with $x$ conjugate to $u_1$, $y$ conjugate to $u_2$, and $(xy)^n=u_i$ for some $n$, and that $C_2$ then has a noncyclic subgroup $V$ such that all involutions in $V$ are conjugate to $u_2$ in $G$. This is part of exercise 10 of page 83, Jacobson's Basic Algebra I. Thanks for any help.
My sparse thoughts: I know a few facts I think are useful (proofs are linked in the numbers to the left hand side):
1. If $c_i=|C(u_i)|$ and $s_i=|S_i|$, then $|G|=c_1s_2+c_2s_1$.
2. $C_2$ is a Sylow $2$-subgroup, and I observe from this that one can assume $u_1\in C_2$ by taking a conjugate.
3. There are $3$ classes of involutions in $C_2$, and if $x$ is an involution distinct from $u_2$, then $x$ is conjugate to $xu_2$ in $C_2$.
I'm lost on how to use this info on the involutions in $C_2$ to count the size of $S_2$. Identifying $C_2$ with $D_8=\langle r,s\mid r^4=s^2=1,\; sr=r^3s\rangle$ and $r^2$ with $u_2$, I know that the two noncyclic subgroups of order $4$ are $\{1,s,r^2,r^2s\}$ and $\{1,rs,r^2,r^3s\}$. I consider the intersection $C:=C_1\cap C_2$. This is a subgroup with order dividing $8$ and $12$, but since I already know $1,u_1,u_2\in C$, then $|C|=4$. I think this might be the desired subgroup $V$. I'm not sure what action of $C_1$ on $u_2$ to consider, since I don't know if left multiplication or conjugation will send $u_2$ back into $C_2$.