If a cyclic group has an element of infinite order, how many elements of finite order does it have?
I know that the order of the entire group must be infinite, for an element of the group must have an order less than the group order. My first thought was that there are no elements with finite order in this group, however now I'm believing that there are infinitely many elements of finite order, since the group should have infinitely many elements. Can someone explain the solution?
Best Answer
Your first thought was almost right: the only element of finite order is the identity. If $G$ is an infinite cyclic group, then $G=\{g^n:n\in\Bbb Z\}$ for some $g\in G$. If $x\in G$, then $x=g^m$ for some $m\in\Bbb Z^+$. If $m\ne 0$, so that $x$ is not the identity element, then for each $k\in\Bbb Z$ we have $x^k=(g^m)^k=g^{km}$, and the $g^{km}=g^0$ if and only if $km=0$ and hence if and only if $k=0$. In other words, no positive power of $x$ is the identity, and therefore $x$ has infinite order.