Subgroup Generated by Infinite Order Element is Infinite

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