# 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 $x \in G \setminus \set e$.

$\order x > 1$

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:

 $\displaystyle \order x$ $=$ $\displaystyle 3$ $\displaystyle \order {x^3}$ $=$ $\displaystyle 3$ $\displaystyle \order {x^9}$ $=$ $\displaystyle 3$

Hence one of the following holds:

 $\displaystyle \order {\gen x}$ $=$ $\displaystyle 3$ $\displaystyle \order {\gen {x^3} }$ $=$ $\displaystyle 3$ $\displaystyle \order {\gen {x^9} }$ $=$ $\displaystyle 3$

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 $3$.

$\blacksquare$