Order of Isomorphic Image of Group Element

Theorem
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$.

Let $\phi: G \to H$ be a group isomorphism.

Then:
 * $a \in G \implies \left|{\phi \left({a}\right)}\right| = \left|{a}\right|$

Proof
First, suppose $a$ is of finite order.

By definition, $\phi$ is bijective, therefore injective.

The result then follows from Order of Homomorphic Image of Group Element.

Now suppose $a$ is of infinite order.

Suppose $\phi \left({a}\right)$ is of finite order.

Consider the mapping $\phi^{-1}: H \to G$.

Let $b = \phi \left({a}\right)$.

Let $\left|{b}\right| = m$.

Then:
 * $\left|{a}\right| = \left|{\phi^{-1} \left({b}\right)}\right| = m$

and that would mean $a$ was of finite order.

The result follows by transposition.