Order of Homomorphic 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 homomorphism.

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

Then:
 * $\forall g \in G: \left|{\phi \left({g}\right)}\right| \backslash \left|{g}\right|$

where $\backslash$ denotes divisibility.

If $\phi$ is injective, then $\left|{\phi \left({g}\right)}\right| = \left|{g}\right|$.

Proof

 * Let $\phi: G \to H$ be a homomorphism.

Let $\left|{g}\right| = n, \left|{\phi \left({g}\right)}\right| = m$.

It follows from Element to the Power of Multiple of Order that $m \backslash n$.


 * Now suppose $\phi: G \to H$ is injective.

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

From the definition of the Order of an Element, that means $n \le m$ since $n$ is the smallest such power.

Thus $m = n$ and the result holds.