Image is a subgroup

Proposition (Image is a subgroup). Let $f:G\to H$ be a group homomorphism . Let $\mathrm{im}(f)$ be its image . Then $\mathrm{im}(f)$ is a subgroup of $H$.

Context. Together with “kernel is normal,” this gives the basic structural picture of any homomorphism: it factors through a quotient of $G$ onto a subgroup of $H$.

Proof sketch. $e_H=f(e_G)\in \mathrm{im}(f)$. If $x=f(g_1)$ and $y=f(g_2)$ lie in the image, then

Apply the one-step subgroup test in $H$.