Let $~$f: G \to H$~$ 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(gh)$~$, 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$~$.