Group of Order 48 is not Simple

Theorem

Let $G$ be of order $48$.

Then $G$ is not simple.

Proof

Aiming for a contradiction, suppose $G$ is simple.

We have that:

$48 = 2^4 \times 3$

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

From Sylow $2$-Subgroups in Group of Order $48$, $n_2$ is either $1$ or $3$.

Let $P$ be a Sylow $2$-subgroup of $G$.

Let $n_2 = 3$.

By Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup, the normalizer of $P$ has index $3$.

By Order of Simple Group divides Factorial of Index of Subgroup:

$\order G \divides 3!$

But $48$ does not divide $6$.

Hence it cannot be the case that $n_2 = 3$.

Hence $n_2 = 1$.

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

This contradicts the assumption that $G$ is simple.

Hence $G$ is not simple.

$\blacksquare$