Group of Order 54 has Normal Subgroup of Order 27

Theorem
Let $G$ be of order $54$.

Then $G$ has a normal subgroup of order $27$.

Proof
We have that:
 * $54 = 2 \times 3^3$

From the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup, which is of order $3^3 = 27$.

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:
 * $n_3 \divides 2$

where $\divides$ denotes divisibility.

It follows that $n_3 = 1$.

From Sylow $p$-Subgroup is Unique iff Normal, this Sylow $3$-subgroup is normal.