# 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 $3$.

$\blacksquare$

## Sources

- 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 44$. Some consequences of Lagrange's Theorem