Group of Order 54 has Normal Subgroup of Order 27
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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.
$\blacksquare$