Let ${\rm Hom}:G\to G'$ be a homomorphism. Prove that if $H\unlhd G$ then ${\rm Hom}(H)\unlhd G'$.

Question

Let ${\rm Hom}:G\to G'$ be a homomorphism. Prove that if $H\unlhd G$ then ${\rm Hom}(H)\unlhd G'$.

My current approach is to show that ${\rm Hom}(gHg^{-1}) = g'{\rm Hom}(H)g'^{-1}$ using the properties of homomorphisms and normal subgroups but it seems too simple for this to be correct.

Answer 1

This is not true. Take any $H$ that is not a normal subgroup of a group $G$.

Then the inclusion of $H$ in $G$ is a group homomorphism and the image of $H$ is obviously not normal in $G$.

This is however true if the homomorphism is supposed to be onto.

Answer 2

Counterexample: consider any example of a non-normal subgroup $G$ of a group $G'$. The identity map $\varphi : G \hookrightarrow G'$ is a homomorphism. $H=G$ is a normal subgroup of $G$, while $\varphi(H) = G$ is a non-normal subgroup of $G$ by assumption.

However, if the homomorphism is surjective, then the claim is true. To show $g' \text{Hom}(H) g'^{-1} = \text{Hom}(H)$ write $g'$ as the image of some element $g$ of $G$ and use your work.