Proving inverse image is a normal subgroup Let $\Phi: G \to G'$ be a group homomorphism, and $H' \lhd G'$. We need to prove that $\Phi^{-1} (H') \lhd G$. 
Here is my current attempt at this. I have already proved that $\Phi^{-1} (H') \leq G$, so the only thing left to show is that $\Phi^{-1} (H')$ is closed under conjugation by $G$. 
By definition, 
$$\Phi^{-1}(H') = \{g \in G' \mid \Phi(g) \in H' \}.$$
Let $a \in \Phi^{-1}(H')$ and $b \in G'$. Consider the element $bab^{-1}$. We will show $bab^{-1} \in \Phi^{-1} (H')$, which is true if and only if $\Phi(bab^{-1}) \in H'$. By the homomorphism property of $\Phi$, we have
$$\Phi(bab^{-1}) = \Phi(b) \Phi(a) \Phi(b^{-1}) = \Phi(b) \Phi(a) \Phi(b)^{-1}.$$
This is where I am stuck. If $\Phi$ were onto, this would be fine. We'd be able to write any element of $G'$ as $\Phi(b)$ for some $b \in G$. We already have $\Phi(a) \in G'$. So this product would live in $H'$ since $H'$ is a normal subgroup. Absent $\Phi$ being onto, how can I finish the proof?
 A: Let $a\in\Phi^{-1}(H')$ and $g\in G$. Note that $\Phi(a)\in H'$ and since  $H'$ is normal $$\Phi (gag^{-1})=\Phi(g) \Phi(a) \Phi(g)^{-1}\in H'.$$ Hence $gag^{-1}\in\Phi^{-1}(H')$.
A: If $a\in \Phi^{-1}(H')$ and $g\in G$ you want to show $g a g^{-1}\in \Phi^{-1}(H')$.
By applying $\Phi$ using the fact that $\Phi$ is homomorphism,we will get that $g ag^{-1}\in \Phi^{-1}(H') if$ $\Phi(g)\Phi(a) \Phi(g)^{-1} \in H'$. 
Now, since $H'$ is normal, $\Phi(g) \Phi(a) \Phi(g)^{-1}\in H'$, because $\Phi(a)\in H'$ (since $a\in \Phi^{-1}(H')$ ).
A: If $H' \lhd G'$, then $\forall a \in G$, it is $\Phi(a)^{-1}H'\Phi(a) \subseteq H'$; but $\Phi(a)^{-1}=\Phi(a^{-1})$, thence:
$$\Phi(a^{-1})H'\Phi(a) \subseteq H', \forall a \in G \tag 1$$
By definition, $H'=\Phi(\Phi^{\leftarrow}(H'))$, so $(1)$ reads:
$$\Phi(a^{-1})\Phi(\Phi^{\leftarrow}(H'))\Phi(a)=\Phi(a^{-1}\Phi^{\leftarrow}(H')a) \subseteq \Phi(\Phi^{\leftarrow}(H')), \forall a \in G \tag 2$$
which implies:
$$a^{-1}\Phi^{\leftarrow}(H')a \subseteq \Phi^{\leftarrow}(H'), \forall a\in G \tag 3$$
namely $\Phi^{\leftarrow}(H') \lhd G$
