Why are there $3$ conjugacy classes of involutions in this centralizer? 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.
 A: The dihedral group of order $8$, $D_8 = \langle r,s \mid r^4 = s^2 = 1, sr = r^3s\rangle$, has exactly five elements of order $2$: $r^2$, which is central, and $s$, $rs$, $r^2s$, and $r^3s$, which are not central.
Since $u_2\in C(u_2)$ and is central (since it commutes with everything in its centralizer), $u_2$ is a conjugacy class by itself, and corresponds to the element $r^2$ in $D_8$.
Now notice that the conjugates of $s$ are just $s$ and $r^2s$: conjugating by a power of $r$ we get $r^isr^{-i} = r^{2i}s$, so we either get $r^2s$ (when $i$ is odd) or $s$ (when $i$ is even). And conjugating by an element of the form $r^is$ we get $(r^is)s(sr^{-i}) = r^isr^{-i} = r^{2i}s$ again.
Similarly, the conjugates of $rs$ are $rs$ and $r^3s$. Now, notice that $r^3s = (rs)r^2$, $rs = (r^3s)r^2$, $s=(r^2s)r^2$, and $r^2s = sr^2$. So, in each case, if $x\in D_8$ is an involution different from $r^2$, then the other conjugate of $x$ is $xr^2$.
So this happens simply because you have a group isomorphic to the dihedral group of order $8$, and nothing to do with the fact that it is a centralizer of an involution of a group with exactly two conjugacy classes of involutions.
Note also that you will not necessarily have $u_1u_2\in C_2$: that would require $c_1$ and $c_2$ to centralize each other, and this will depend on the representatives you choose (since, in $S_5$, the two classes of involutions are the transpositions $(i,j)$, and the products of two transpositions $(r,s)(t,u)$; these may commute, e.g., $(1,2)$ commutes with $(1,2)(3,4)$; but they need not commute, as for example $(1,2)$ does not commute with $(1,3)(4,5)$ ).
