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

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Theorem

Let $G$ be a non-abelian finite simple group.

Let $t \in G$ be a self-inverse element of $G$.


Then:

$\map {C_G} t \ne G$

where $\map {C_G} t$ denotes the centralizer of $t$ in $G$.


Proof

Let $G$ be a non-abelian finite simple group.

Let $t \in G$ which is not the identity.


By definition of a simple group and Center of Group is Normal Subgroup:

either $\map Z G = G$ or $\map Z G$ is the trivial group.

By definition of an abelian group:

$\map Z G = G$ if and only if $G$ is abelian

Hence we must have $\map Z G$ is the trivial group.

Thus $t \notin \map Z G$.


From definition of center:

$\exists x \in G: t x \ne x t$

For this $x$, $x \notin \map {C_G} t$.

Hence $\map {C_G} t \ne G$.


The validity of the material on this page is questionable.
In particular: So trivial it's scary: something stronger was proved
Remark: the identity is a self-inverse element with $\map {C_G} e = G$ trivially
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