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:
\(\ds \order x\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds \order {x^3}\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds \order {x^9}\) | \(=\) | \(\ds 3\) |
Hence one of the following holds:
\(\ds \order {\gen x}\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds \order {\gen {x^3} }\) | \(=\) | \(\ds 3\) | ||||||||||||
\(\ds \order {\gen {x^9} }\) | \(=\) | \(\ds 3\) |
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