If $\Phi: G \to G'$ is onto and $H \lhd G$, then $\Phi(H) \lhd G'$. Let $\Phi: G \to G'$ be an onto group homomorphism and $H \lhd G$. I need to prove that $\Phi(H) \lhd G'$. 
I have already been able to prove that $\Phi(H)$ is a subgroup of $G'$, so all that remains to show is that it is closed under conjugate by elements of $G'$. 
Let $\alpha \in \Phi(H)$ and $\beta \in G'$, and consider $\beta \alpha \beta^{-1}$, which we must show lives in $\Phi(H)$. Since $\alpha \in \Phi(H)$, by definition there exists $h \in H$ such that $\Phi(h) = \alpha$. Since $\Phi$ is onto, there exists $g \in G$ such that $\Phi(g) = \beta$. This implies that $\Phi(g)^{-1} = \beta^{-1}$ since inverses are unique and, by the properties of homomorphism, that $\Phi(g^{-1}) = \beta^{-1}$. By the homomorphism property, we get: 
\begin{align*}
\beta \alpha \beta^{-1} = \Phi(g) \Phi(h) \Phi(g^{-1}) = \Phi(ghg^{-1}). 
\end{align*}
Since $H \lhd G$,  it is closed under conjugation by elements of $G$, so $ghg^{-1} \in H$. Hence, $\exists h = ghg^{-1} \in H$ such that $\Phi(h) = \beta \alpha \beta^{-1}$, so $\beta \alpha \beta^{-1} \in \Phi(H)$. Hence, $\Phi(H) \lhd G'$. 
How does this look? 

We have to prove that for every $g' \in G'$ it is $g'^{-1}\Phi(H)g' \subseteq \Phi(H)$. Since the homomorphism is onto, for every $g' \in G'$, it does exist $g \in G$ such that $g'=\Phi(g)$. But then, $g'^{-1}\Phi(H)g'=\Phi(g)^{-1}\Phi(H)\Phi(g)=\Phi(g^{-1})\Phi(H)\Phi(g)=\Phi(g^{-1}Hg)\subseteq \Phi(H)$, because $g^{-1}Hg \subseteq H$ by hypothesis.
