# Finite group having at least two conjugacy classes of involutions has order less than the cube of the maximum of orders of centralizers of involutions

## Statement

Suppose $G$ is a finite group such that there are at least two conjugacy classes of involutions (non-identity elements of order two) in $G$. Then, if $h$ is the maximum of the orders of all subgroups of $G$ that arise as a Centralizer of involution (?), we have: $|G| < h^3$.

## Proof

Given: A group $G$ with at least two conjugacy classes of involutions. $h$ is the maximum possible order of a subgroup arising as the centralizer of an involution of $G$.

To prove: $|G| < h^3$.

Proof:

1. Let $x$ be an involution of $G$ such that $|C_G(x)|$ has order $h$ and let $C = C_G(x)$. Note that $x$ exists by the definition of $h$.
2. Let $y$ be an involution of $G$ that is not conjugate to $x$ and let $D = C_g(y)$: Note that $y$ exists because of the assumption that there are at least two conjugacy classes of involutions.
3. Let $y = y_1, y_2, \dots, y_t$ be distinct involutions of $D$, and define $D_i = C_G(y_i)$. In particular, $D_1 = D$.
4. $t \le |D| \le h$: $t \le |D|$ because the $y_i$ are all distinct elements of $D$. $|D| \le h$ because $D$ is a centralizer of involution and by definition, $h$ is the maximum of the orders of such subgroups.
5. The number of distinct non-identity elements in $\bigcup_{i=1}^t D_i$ is less than $h^2$: Each $D_i$ has at most $h - 1$ non-identity elements, and there are $t$ of them, with $t \le h$. So, the total number of elements is less than $h^2$.
6. Suppose $x$ has a total of $m$ conjugates $x_1, x_2, \dots, x_m$. Then, each $x_j$ is contained in the union $\bigcup_{i=1}^t D_i$: Any $x_j$ is conjugate to $x$. Since by assumption $x$ is not conjugate to $y$, $x_j$ is not conjugate to $y$. Thus, by fact (1), there exists an involution centralizing both $x_j$ and $y$. This involution lives in $C_G(y)$ so it is some $y_i$. Thus, $x_j \in C_G(y_i)$ for some $i$.
7. $m < h^2$: This follows from the previous two steps.
8. $|G| < h^3$: First, note that under the action of the group on itself by conjugation, the centralizer of $x$ is its stabilizer, so by fact (3), the coset space of $C_G(x)$ in $G$ is in bijection with the conjugacy class of $x$. Thus, $m = [G:C_G(x)]$. By fact (4) (Lagrange's theorem), we get $m = |G|/h$. Since $m < h^2$ by the previous step, we get $|G| < h^3$.