Group of Order 27 has Subgroup of Order 3

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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:

\(\displaystyle \order x\) \(=\) \(\displaystyle 3\)
\(\displaystyle \order {x^3}\) \(=\) \(\displaystyle 3\)
\(\displaystyle \order {x^9}\) \(=\) \(\displaystyle 3\)


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 $3$.

$\blacksquare$


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