Talk:Centralizer of Self-Inverse Element of Non-Abelian Finite Simple Group is not That Group

In 1978: John S. Rose: A Course on Group Theory: $1$: Introduction to Finite Group Theory: $1.14$, the result is posed as:

(R. Brauer and K. A. Fowler [a8], 1955) Let $G$ be a non-abelian finite simple group (so that, by $1.12$, $\order G$ is even) and let $t$ be an involution in $G$. Then $\map {C_G} t \ne G$, and if $\order {\map {C_G} t} = m$ then $\order G \le \paren {\frac 1 2 m \paren {m + 1} }!$

where reference [a8] is the one cited.

In $1.13$ he states:

Let $G$ be any group of even order. Then $G$ possesses at least one element of order $2$. (Any such element is called an involution.)

This gives the definition of involution.

I have not inspected Brauer and Fowler to see what it says, but that looks like the next stop along the way. Feel free to take this up. --prime mover (talk) 10:07, 20 May 2020 (EDT)

I see. In that case $1.13$ is proved in Even Order Group has Order 2 Element, and $1.14$ was broken down to two theorems.

I think I'll make a new page for the general result, and the result for involutions as a direct consequence.

However a point still stands: An involution defined in Rose (order 2 element) does not include the identity, but Definition:Self-Inverse Element does, and they are both reasonably defined.

Brauer and Fowler gives this relevant result in Corollary 2I, $7$ pages in their paper:

If $G$ is a group of even order $g > 2$ which contains $m$ involutions and if $n = g / m$, then

$g < \sqbrk {n \paren {n + 1} / 2}!$

They also proved $n \le \order {\map {N_G} t}$ for an involution $t$, not sure how it ties in to their centralizers though.

--RandomUndergrad (talk) 12:52, 20 May 2020 (EDT)