Subgroup Generated by Infinite Order Element is Infinite
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Theorem
Let $G$ be a group.
Let $a \in G$ be of infinite order.
Let $\gen a$ be the subgroup generated by $a$.
Then $\gen a$ is of infinite order.
Proof
Aiming for a contradiction, suppose $\gen a$ is of finite order.
We have that $a \in \gen a$ by definition.
From Element of Finite Group is of Finite Order it follows that $a$ is of finite order.
From this contradiction it follows that $\gen a$ must be of infinite order after all.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 38.2$ Period of an element