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

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.