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The following is an idea originally communicated by Richard Brauer. I'm having difficulty following some of the combinatorial elements.

Let $G$ be a finite group containing exactly two conjugacy classes of involutions. Let $u_1$ and $u_2$ be nonconjugate involutions in $G$. Let $c_i=|C(u_i)|$, the respective centralizers of the $u_i$. Let $S_i$ be the set of ordered pairs $(x,y)$ with $x$ conjugate to $u_1$, $y$ conjugate to $u_2$, and $(xy)^n=u_i$ for some $n$. Let $s_i=|S_i|$. Then $|G|=c_1s_2+c_2s_1$.

The idea is to count the ordered pairs $(x,y)$ with $x$ conjugate to $u_1$ and $y$ conjugate to $u_2$ in two different ways. Is it somehow clear that this number is $(|G|/c_1)(|G|/c_2)$ right off the bat? I believe this follows since the possible choices of the $x$ is the cardinality of the orbit of $u_1$ under the conjugation action, which equals the index of the stabilizer in $G$, but the stabilizer is just the centralizer in this case. Likewise for the choices of $y$.

On the other hand, I know that if the product $uv$ of two involutions $u$ and $v$ has odd order, then $u$ and $v$ are conjugate. So if $x$ and $y$ are not conjugate here, their product $xy$ has even order. Then $(xy)^n$ for $n=o(xy)/2$ is an involution, and thus conjugate to either $u_1$ or $u_2$. This exhausts all the pairs in question, but why does counting in this way give $(|G|/c_1)s_1+(|G|/c_2)s_2$? I only know that $(xy)^n$ is conjugate to some $u_i$ for some $n$, not that $(xy)^n=u_i$ as in the definition of the $s_i$.

After that, the result will follow from the equality $$ (|G|/c_1)(|G|/c_2)=(|G|/c_1)s_1+(|G|/c_2)s_2. $$

Thanks.

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up vote 2 down vote accepted

$G$ acts on the conjugacy class of $u_i$ by conjugation, so the number of elements conjugate to $u_i$ is the index of the stabilizer of $u_i$ in $G$. The stabilizer is $C(u_i)$, so the number of involutions conjugate to $u_i$ is $|G|/c_i$.

For your second question:

Let $S$ be the set of pairs $(x,y)$ such that $x$ is conjugate to $u_1$ and $y$ is conjugate to $u_2$.

Let $T_i$ be the set of pairs $(x,y) \in S$ such that $(xy)^n$ (for $n=o(xy)/2$) is conjugate to $u_i$.

You want to prove that $|T_i| = |S_i| (|G|/c_i)$.

For each involution $u$, let $S_u$ be the set of pairs $(x,y) \in S$ such that $(xy)^n$ (for $n=o(xy)/2$) equals $u$.

We then have $|T_i| = \sum |S_u|$, where the sum takes place over the conjugacy class of $u_i$. But the size of $S_u$ is independent of the choice of $u$ within the conjugacy class of $u_i$, because if $u^g = v$ then $(x,y) \mapsto (x^g, y^g)$ is a bijection between $S_u$ and $S_v$. Hence $|S_u| = |S_i|$ for all $u$ in the conjugacy class of $u_i$. There are $|G|/c_i$ conjugates of $u_i$, so $|T_i| = |S_i| (|G|/c_i)$.

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Yes. I have fixed it. –  Ted Jun 5 '12 at 4:41
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