Group of Order 27 has Subgroup of Order 3

Theorem
Let $G$ be a group whose identity element is $e$.

Let $G$ be of order $27$.

Then $G$ has at least one subgroup of order $3$.

Proof
Let $x \in G \setminus \set e$.

From Identity is Only Group Element of Order 1:
 * $\order x > 1$

where $\order x$ denotes the order of $x$.

From Lagrange's Theorem, $\order x$ is $3$, $9$ or $27$.

Thus one of the following applies:

Hence one of the following holds:

where $\gen x$ denotes the subgroup generated by $x$.

Thus one of these: $\gen x$, $\gen {x^3}$ or $\gen {x^9}$ is a subgroup of $G$ of order $3$.