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:

$a \in G \implies \order {\map \phi a} = \order a$

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.

$\Box$

Now suppose $a$ is of infinite order.

Suppose $\map \phi a$ is of finite order.

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

Let $b = \map \phi a$.

Let $\order b = m$.

Then:

$\order a = \order {\map {\phi^{-1} } b} = m$

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

The result follows by transposition.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.3$. Isomorphism: Example $138$