Prove that in an infinite cyclic group order of every element ($\ne e$) is infinite.
I have tried proving this way : If there exists an element of finite order, then it must generate a finite subgroup of $G.$ But this is not possible as by Lagrange's theorem, order of a subgroup must divide order of a group, which is not true here as order of $G$ is infinity.
Is this correct and complete ?
Best Answer
This proof is invalid, since some infinite groups have finite subgroups. E.g. $\mathbb{Z}_2 \times \mathbb{Z}$ has the subgroup $\mathbb{Z}_2 \times \{0\}$.
In an infinite cyclic group the elements are $$\{\ldots,g^{-3},g^{-2},g^{-1},1,g,g^2,g^3,\ldots,\}$$ for some generator $g$.
If $h \neq 1$ is in this group, then $h=g^k$ for some $k \in \mathbb{Z} \setminus \{0\}$. So, the subgroup generated by $h$ has the underlying set $$\{\ldots,g^{-3k},g^{-2k},g^{-k},1,g,g^{2k},g^{3k},\ldots,\}.$$