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:

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

where $\divides$ denotes divisibility.

Proof

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

Let $\order g = n, \order {\map \phi g} = m$.

\(\ds \paren {\map \phi g}^n\)

\(=\)

\(\ds \map \phi {g^n}\)

Homomorphism of Power of Group Element

\(\ds \)

\(=\)

\(\ds \map \phi {e_G}\)

\(\ds \)

\(=\)

\(\ds e_H\)

Homomorphism to Group Preserves Identity

It follows from Element to Power of Multiple of Order is Identity that $m \divides n$.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $4 \ \text{(ii)}$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Exercise $4$