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 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 $8$.

The post Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8 first appeared on Problems in Mathematics.