Isomorphisms preserve order: the infinite case Given a group $G$ and a isomorphism $\varphi: G \to H$, I know how to prove that $|x| = |\varphi(x)|$ in the finite case. If $|x| = n$, I prove that $\varphi(x^n) = \varphi(x)^n$, and since $x^n = e$,  $\varphi(x^n) = \varphi(e_G) = e_H$, so $\varphi(x)^n = e_H$, so $|\varphi(x)| \mid |n|$. On the flip side, let $m = |\varphi(x)|$. Then $\varphi(x)^m = e_H$. But then
$$
e_H = \varphi(x)^m = \varphi(x^m).
$$
As $\varphi$ is an isomorphism, it is injective; since $\varphi(x^m) = e_H = \varphi(e_G)$, we have $x^m = e_G$, so $|x| \mid |\varphi(x)|$, so $|x|=|\varphi(x)|$.
This proof does not generalize to the case where $|x|$ or $|\varphi(x)|$ are allowed to be infinite, however. I can't figure out how to prove the statement in that case. I could argue that the above shows that $|x|$ is finite if and only if $|\varphi(x)|$ is finite, so one is infinite if and only if the other is infinite. I don't think that proves the statement in general, however.
 A: You might find it useful to prove that the order of the element $x\in G$, where $G$ is a group, is the cardinality (finite or infinite) of $\langle x\rangle$, the cyclic subgroup generated by $x$.
An isomorphism $\varphi\colon G\to H$ is a bijective map (so it preserves the cardinality of subsets) and $\varphi(\langle x\rangle)=\langle\varphi(x)\rangle$.
How do you prove the first statement? Consider the homomorphism $f_x\colon\mathbb{Z}\to G$ defined by $f_x(k)=x^k$. The image of $f_x$ is precisely $\langle x\rangle$.
If $x$ has infinite order, then $f_x$ is injective, so $\langle x\rangle$ is infinite.
Suppose $x$ has finite order $n$. Then, by definition, $x^n=e$ and for no integer $k$ with $0<k<n$ we have $x^k=e$.
The kernel of $f_x$ is a subgroup of $\mathbb{Z}$, so it is of the form $m\mathbb{Z}$ for some $m>0$ and the minimum positive integer in $m\mathbb{Z}$ is $n$ by assumption. Therefore $m=n$.
From the homomorphism theorem it follows that $\langle x\rangle\cong\mathbb{Z}/\ker f_x=\mathbb{Z}/n\mathbb{Z}$, which has $n$ elements.
