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 $S_{X}$.

Subgroup of Finite Index Contains a Normal Subgroup of Finite Index
Let $G$ be a group and let $H$ be a subgroup of finite index. Then show that there exists a normal subgroup $N$ of $G$ such that $N$ is of finite index in $G$ and $N\subset H$.
Proof.
The group $G$ acts on the set of left cosets $G/H$ by left multiplication.
Hence […]

A Subgroup of the Smallest Prime Divisor Index of a Group is Normal
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 […]

Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8
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 $|P|=8$.
Consider the action of the group $G$ on […]

A Group Homomorphism is Injective if and only if the Kernel is Trivial
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 […]

Surjective Group Homomorphism to $\Z$ and Direct Product of Abelian Groups
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 […]

A Group Homomorphism is Injective if and only if Monic
Let $f:G\to G'$ be a group homomorphism. We say that $f$ is monic whenever we have $fg_1=fg_2$, where $g_1:K\to G$ and $g_2:K \to G$ are group homomorphisms for some group $K$, we have $g_1=g_2$.
Then prove that a group homomorphism $f: G \to G'$ is injective if and only if it is […]

Group Homomorphism, Preimage, and Product of Groups
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 $f(g)\in f(H)$.
It follows that there exists $h\in H$ […]

Any Finite Group Has a Composition Series
Let $G$ be a finite group. Then show that $G$ has a composition series.
Proof.
We prove the statement by induction on the order $|G|=n$ of the finite group.
When $n=1$, this is trivial.
Suppose that any finite group of order less than $n$ has a composition […]