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 the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
\[\phi: G\to S_{G/P},\]
where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.

This homomorphism is defined by
\[\phi(g)(aP)=gaP\]
for $g\in G$ and $aP\in G/P$.

Then by the first isomorphism theorem, we see that
\[G/\ker(\phi) \cong \im(\phi) < S_{G/P}.\]
This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$.
Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$.
Thus, we must have $4\mid |\ker{\phi}|$.

Also note that $\ker(\phi) < P$.
To see this let $x\in \ker(\phi)$.
Then we have
\[xP=\phi(x)(P)=\id(P)=P.\]
Here $\id$ is the identity map from $G/P$ to itself.
Hence $x\in P$.
It follows that $|\ker(\phi)|$ divides $|P|=8$.

Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.

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.
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Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$.
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(Michigan State University, Abstract Algebra Qualifying […]

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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 $|X|!$.
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Let $G$ be a finite group of order $21$ and let $K$ be a finite group of order $49$.
Suppose that $G$ does not have a normal subgroup of order $3$.
Then determine all group homomorphisms from $G$ to $K$.
Proof.
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Let $G$ be a finite group. Then show that $G$ has a composition series.
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We prove the statement by induction on the order $|G|=n$ of the finite group.
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Let $G$ be a finite group of order $n$ and suppose that $p$ is the smallest prime number dividing $n$.
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Hint.
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Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.
Hint.
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