Order of Monomorphic Image of Group Element

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

Let $\phi: G \to H$ be a monomorphism.

Let $g \in G$ be of finite order.

Then:
 * $\forall g \in G: \order {\map \phi g} = \order g$

Proof
By definition of monomorphism, $\phi$ is a homomorphism which is also an injection.

From Order of Homomorphic Image of Group Element:
 * $\forall g \in G: \order {\map \phi g} \divides \order g$

So $g^m = e$, as $\phi$ is injective.

From the definition of order of group element, that means $n \le m$ since $n$ is the smallest such power.

Thus $m = n$ and the result holds.