# Nontrivial Action of a Simple Group on a Finite Set

## Problem 112

Let $G$ be a simple group and let $X$ be a finite set.
Suppose $G$ acts nontrivially on $X$. That is, there exist $g\in G$ and $x \in X$ such that $g\cdot x \neq x$.
Then show that $G$ is a finite group and the order of $G$ divides $|X|!$.

## Proof.

Since $G$ acts on $X$, it induces a permutation representation
$\rho: G \to S_{X}.$

Let $N=\ker(\rho)$ be the kernel of $\rho$.
Since a kernel is normal in $G$ and $G$ is simple, we have either $N=\{e\}$ or $N=G$.

If $N=G$, then for any $g\in G$ we have $\rho(g)$ is a trivial action, that is, $g\cdot x=x$ for any $X$.
This contradicts the assumption that $G$ acts nontrivially on $X$.
Hence we have $N=\{e\}$, and it follows that the homomorphism $\rho$ is injective.

Thus we have
$G \cong \mathrm{im} (\rho) < S_{X}.$ Since $S_{X}$ is a finite group and $G$ is isomorphic to its subgroup, the group $G$ is finite.
By Lagrange’s theorem, the order $|G|=|\mathrm{im}(\rho)|$ of $G$ divides the order $|S_{X}|=|X|!$ of $S_{X}$.

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