Is my proof ok?
Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\leq G$. $\text{Im}(H) = \{f(x):x\in H\}$. To show that $\text{Im}(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in \text{Im}(H)$.
$$f(x)f(y)^{-1}=f(xy^{-1})\in \text{Im}(H)\text{ because } xy^{-1}\in H$$
$\text{Ker}(f)=\{x:f(x)=e^{\prime}\}$ ($e^{\prime}$ is the identity in $G^{\prime}$).
$$f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in \text{Ker}(f)$$.
Best Answer
It's ok but don't forget to show that $\mathrm{Im}(H)\ne \emptyset$ since $f(e)=e'\in\mathrm{Im}(H)$ and the same thing for $\ker f$.