# Every Finite Group is a torsion group?

I am learning about finitely generated groups at the moment and in the text book, they mention torsion groups, which they give the definition as:

A group $G$ is a torsion group if every element of $G$ is of finite order.

They then go on to state, when listing examples of torsion groups, that

"Every finite group is a torsion group"

I understand how cyclic groups would be a torsion group, but I am having trouble understand how a finite permutation group would be a torsion group. If anyone could help clarify why, it would be greatly appreciated.

• Every permutation of a finite set has finite order. If a group is finite the sequence of powers of any element must repeat, so the element has finite order. – Ethan Bolker Jun 21 '17 at 15:03
• Suppose $x \in G$ has not finite order: then the subgroup generated by $x$ is infinite. This shows that $G$ has infinitely many elements. Hence, if you consider a finite group $G$. necessarily all of its elements have finite order. – Crostul Jun 21 '17 at 15:03
• Thank you both for your comments, it makes sense now. – Smeef Jun 21 '17 at 15:07

The order of an element $g$ is the cardinality of $\langle g \rangle$, the cyclic group generated by it.
Every element $g$ in a finite group $G$ has finite order because $\langle g \rangle \subseteq G$ implies that $\langle g \rangle$ is finite.