Order of Monomorphic Image of Group Element

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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:

$\forall g \in G: \order {\map \phi g} \divides \order g$


\(\ds \map \phi {g^m}\) \(=\) \(\ds \paren {\map \phi g}^m\) Homomorphism of Power of Group Element
\(\ds \) \(=\) \(\ds e_H\)
\(\ds \) \(=\) \(\ds \map \phi {e_G}\) Homomorphism to Group Preserves Identity

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.

$\blacksquare$


Sources

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