Group of Order 36 is not Simple

Theorem

Let $G$ be of order $36$.

Then $G$ is not simple.

Proof

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

We have that:

$36 = 2^2 \times 3^2$

Let $n_3$ denote the number of Sylow $3$-subgroups of $G$.

From the Fourth Sylow Theorem:

$n_3 \equiv 1 \pmod 3$

and from the Fifth Sylow Theorem:

$n_3 \divides 4$

where $\divides$ denotes divisibility.

Hence $n_3$ must be either $1$ or $4$.

Let $P$ be a Sylow $3$-subgroup of $G$.

Let $n_3 = 4$.

By Number of Sylow p-Subgroups is Index of Normalizer of Sylow p-Subgroup, the normalizer of $P$ has index $4$.

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

$\order G \divides 4!$

But $36$ does not divide $24$.

Hence it cannot be the case that $n_3 = 4$.

Hence $n_3 = 1$.

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

This contradicts the assumption that $G$ is simple.

Hence $G$ is not simple.

$\blacksquare$