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:

$\left|G\right|<h^3$.

Related facts

• Dihedral trick
• Every finite group of even order has a proper subgroup of order greater than the cuberoot of the order
• Finite simple non-abelian group has order greater than product of order of proper subgroup and its centralizer
• Finite group having exactly one conjugacy class of involutions need not have order less than the cube of the order of the centralizer of involution

Facts used

1. Involutions are either conjugate or have an involution centralizing both of them
2. Group acts as automorphisms by conjugation
3. Fundamental theorem of group actions

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: $\left|G\right|<h^3$.

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. $t\le \left|D\right|\le h$: $t\le \left|D\right|$ because the $y_i$ are all distinct elements of $D$. $\left|D\right|\le h$ 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. $\left|G\right|<h^3$: 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 $m=\left|G\right|/h$. Since $m<h^2$ by the previous step, we get $\left|G\right|<h^3$.

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 301, Chapter 9 (Groups of even order), Theorem 1.3, ^{More info}