Let be a group homomorphism, and write $f (G)$ for the set ${f (g) : g \in G}$ . Then $f (G)$ is a group under the operation inherited from $H$ .

Proof

To prove this, we must verify the group axioms. Let $f : G \to H$ be a group homomorphism, and let $e_{G}, e_{H}$ be the identities of $G$ and of $H$ respectively. Write $f (G)$ for the image of $G$ .

Then $f (G)$ is closed under the operation of $H$ : since $f (g) f (h) = f (g h)$ , so the result of $H$ -multiplying two elements of $f (G)$ is also in $f (G)$ .

$e_{H}$ is the identity for $f (G)$ : it is $f (e_{G})$ , so it does lie in the image, while it acts as the identity because $f (e_{G}) f (g) = f (e_{G} g) = f (g)$ , and likewise for multiplication on the right.

Inverses exist, by "the inverse of the image is the image of the inverse".

The operation remains associative: this is inherited from $H$ .

Therefore, $f (G)$ is a group, and indeed is a subgroup of $H$ .

The image of a group under a homomorphism is a subgroup of the codomain