Proving cyclic subgroup of a finite group is finite

Question

Prove that if $G$ is finite then $\forall a\in G$, $H:=\langle a\rangle$ is finite.

Pf: Let $G$ be a finite group with order $m$ and let $a \in G$. Suppose that $\langle a \rangle$ is infinite then we see that: $$\langle a\rangle \subset G$$

A contradiction! Since $\langle a \rangle$ is infinite while G is finite.

Any mistake? Thanks in advance!

Answer 1

Your proof is fine.

Alternatively, $H:=\langle a\rangle$ is a subgroup of $G$, so, by Lagrange's Theorem, the order of $H$ divides the order of $G$, so is strictly less than or equal to the order of $G$. But $|G|$ is finite, so $|H|$ is finite.