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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Ethan Bolker Jun 21 '17 at 15:03
  • 2
    $\begingroup$ 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. $\endgroup$ – Crostul Jun 21 '17 at 15:03
  • $\begingroup$ Thank you both for your comments, it makes sense now. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.