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 $$a \in G \Longrightarrow \left|{\phi \left({a}\right)}\right| = \left|{a}\right|$$.

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.