Let $\phi$ be a homomorphism from $G$ to $G'$. Let $H'<G'$.
Show the set $H=\{g\in G \ | \ \phi(g) \in H'\}$ is a subgroup of $G$. (I have done this, but its part 1 of the question)
Prove that if $H'$ is a normal subgroup of $G'$, then $H$ is a normal subgroup of $G$. I know that you have to show that for any $g$ in $G$ that $ghg^{-1}=h$ or $gH=Hg$. I dont see how the normality of $H'$ and $G'$ help in this case.
On
"Intuitive" explanation: Normal subgroups and kernels are the same thing. Any kernel is normal and any normal subgroup $N \leq G$ is the kernel of the quotient map $\pi :G \to G/N$.
A kernel of a map $\phi : H \to G$ is by definition $\phi^{-1}\{0\}$.
Now suppose $N \leq G$ is a normal subgroup.
$\phi^{-1}(N) = \phi^{-1}(\pi^{-1}\{0\})$ = $(\pi \circ \phi)^{-1}\{0\}$
Therefore $\phi^{-1}(N)$ is a kernel, and so normal.
Let $h\in H$ and $g\in G$. We want to show that $ghg^{-1}\in H$. So let's compute $\phi(ghg^{-1})$:
$$ \phi(g h g^{-1}) = \phi(g) \cdot \phi(h) \cdot \phi(g^{-1}) =\phi(g) \cdot \underbrace{\phi(h)}_{\in H'} \cdot \phi(g)^{-1} \in H' $$ because $H'$ is normal in $G'$. But this precisely shows that $ghg^{-1}\in H$.