Normal Subgroup of Group of Order 24

Theorem
Let $G$ be a group of order $24$.

Then $G$ has either:
 * a normal subgroup of order $8$

or:
 * a normal subgroup of order $4$.

Proof
We note that:
 * $24 = 3 \times 2^3$

Hence a Sylow $2$-subgroup of $G$ is of order $8$.

Let $n_2$ denote the number of Sylow $2$-subgroups of $G$.

By the Fourth Sylow Theorem:
 * $n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd

and from the Fifth Sylow Theorem:
 * $n_2 \divides 24$

where $\divides$ denotes divisibility.

It follows that $n_3 \in \set {1, 3}$.

Suppose $n_2 = 1$.

Let $P$ denote the unique Sylow $2$-subgroup of $G$.

From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.

This is therefore the required normal subgroup of order $8$.

Suppose $n_2 = 3$.

Let $S_1$, $S_2$ and $S_3$ denote the $3$ Sylow $2$-subgroups of $G$.

Each has order $8$, as noted earlier.

Consider the product $S_1 S_2$.

By Order of Subgroup Product:
 * $\order {S_1 S_2} = \dfrac {2^3 2^3} {2^r}$

where:
 * $\order {S_1 S_2}$ denotes the order of $S_1 S_2$
 * $\order {S_1 \cap S_2} = 2^r$

As $S_1 S_2 \subseteq G$ and $\order G = 24$:
 * $\order {S_1 S_2} \le 24$

Hence:

As $S_1 \cap S_2$ is a proper subgroup of $S_1$, it can have no more than $2^2 = 4$ elements.

So if $G$ has $3$ Sylow $2$-subgroups, the intersection of any $2$ of them is of order $4$.

Let $T = S_1 \cap S_2$, so that $\order T = 4$.

We have that:
 * $\index T {S_1} = 2$

where $\index T {S_1}$ denotes the index of $T$ in $S_1$.

From Subgroup of Index 2 is Normal, $T$ is normal in $S_1$.

Similarly, $T$ is normal in $S_2$.

Thus $S_1$ and $S_2$ are both subgroups of the normalizer $\map {N_G} T$ of $T$ in $G$.

Thus $H = \gen {S_1 S_2}$ is a subgroup of $\map {N_G} T$.

Hence $T$ is a normal subgroup of $H$.

As $H$ is a subgroup of $\map {N_G} T$, it contains $S_1 S_2$ as a subset.

But $S_1 S_2$ contains $\dfrac {2^6} {2^2} = 16$ elements.

The only subgroup of $G$ containing $16$ elements has to be $G$ itself.

So $H = G$ and so $\map {N_G} T = G$.

That is, $T$ is a normal subgroup of $G$.

This is therefore the required normal subgroup of order $4$.