Endomorphisms of a finite simple group

Statement

Let $G$ be a finite simple group. Every group endomorphism $f:G\to G$ is either the trivial homomorphism $0_G:g\mapsto e$ , or an automorphism. Consequently,

\operatorname{End}_{\mathbf{Grp}}(G)=\{0_G\}\sqcup\operatorname{Aut}_{\mathbf{Grp}}(G), \qquad |\operatorname{End}_{\mathbf{Grp}}(G)|=1+|\operatorname{Aut}_{\mathbf{Grp}}(G)|.

This is a decomposition of a monoid into its zero element and its group of units.

Proof

The kernel is normal, so simplicity gives $\ker f=G$ or $\ker f=\{e\}$ . The first case is the trivial map. In the second case $f$ is injective, hence bijective because its domain and codomain are the same finite set. Its inverse preserves multiplication, so $f$ is an automorphism. Nontriviality of $G$ makes the two cases disjoint.

The role of the category

The same formula holds in the full subcategories of finite groups and finite simple groups. It does not describe arbitrary functions on the underlying set, of which there are $|G|^{|G|}$ .

For two different finite simple groups, a nontrivial homomorphism is injective but need not be surjective. The endomorphism conclusion uses the fact that the source and target coincide.