Let $f:G\to H$ be a group homomorphism, and write $f\left(G\right)$ for the set $\left\{f\right(g):g\in G\}$. Then $f\left(G\right)$ 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 $f\left(G\right)$ for the image of $G$.

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

$e_H$ is the identity for $f\left(G\right)$: it is $f\left(e_G\right)$, so it does lie in the image, while it acts as the identity because $f\left(e_G\right)f\left(g\right)=f\left(e_{G}g\right)=f\left(g\right)$, 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, $f\left(G\right)$ is a group, and indeed is a subgroup of $H$.