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 $|X|!$.

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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Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
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Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.
Then prove that any subgroup of index $p$ is a normal subgroup of $G$.
Hint.
Consider the action of the group $G$ on the left cosets $G/H$ by left […]

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Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.
Proof.
Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.
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Let $G$ be an abelian group and let $f: G\to \Z$ be a surjective group homomorphism.
Prove that we have an isomorphism of groups:
\[G \cong \ker(f)\times \Z.\]
Proof.
Since $f:G\to \Z$ is surjective, there exists an element $a\in G$ such […]

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Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$.
Definitions/Hint.
We recall several […]

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Let $G, G'$ be groups and let $f:G \to G'$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
\[f^{-1}(f(H))=HN.\]
Proof.
$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
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Let $G$ be a finite group. Then show that $G$ has a composition series.
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