Images and preimages of groups using group homomorphism I'm trying to solve this problem from my group theory course:

Let $\phi:G_1\rightarrow G_2$ be a group homomorphism. Also let
$N\unlhd G_1$ (normal subgroup) and $K\leq G_2$.
a) Is $\phi(N)$ a normal subgroup of $G_2$?
b) Is $\phi^{-1}(K)=\{x\in G_1:\phi(x)\in K\}$ a subgroup of $G_1$?
c) If $K$ is normal, is $\phi^{-1}(K)$ a normal subgroup of $G_1$?

The work I've done so far:
a) Bein $y\in G_2$, if $\phi(N)$ is normal in $G_2$ then it must verify $y\phi(N)y^{-1}= \phi(N)$. I've seen that, if $\phi$ is surjective, then $\exists x\in G_1$ so that $\phi(x)=y$, so we can see that
$$y\phi(N)y^{-1}=\phi(x)\phi(N)\phi(x)^{-1}=\phi(xNx^{-1})=\phi(N).$$
So I've proven $\phi(N)\unlhd G_2$ just in the case $\phi$ is surjective (I'm not sure if I need to prove that $\phi$ being non-surjective implies $\phi(N)$ being not normal, or if it is trivial).
b) For this second question, I've seen that, given $x\in G_1$ so that $\phi(x)\in K$, then $\phi(x)^{-1}=\phi(x^-1)\in K$ (since $K$ is group). Also that $\phi(e_1)=e_2\in K$. It obviously verifies associativity, and, given two $x,y\in\phi^{-1}(k)$, then also $\phi(xy)=\phi(x)\phi(y)\in K$, so I conclude that $\phi^{-1}(K)$ is subgroup of $G_1$.
c) We have that, for any $g\in G_1$, (assuming of course $K$ normal subgroup of $G_2$), then
$$\phi(g)K\phi(g^{-1})=K$$
We apply $\phi^{-1}$ to the expression above and we get that
$$\phi^{-1}\left(\phi(g)K\phi(g^{-1})\right)=g\phi^{-1}(K)g^{-1}=\phi^{-1}(K),$$
so we conclude that $K\unlhd G_2$ implies $\phi^{-1}(K)\unlhd G_1$ (Is this statement true even when $\phi$ is not surjective? I think it is but I may be wrong).
Are these solutions correct? If not, why? Any help will be appreciated, thanks in advance.
 A: Statement (a) is false, but you can't prove that $\phi(N)$ is not normal: it may or may not be. How do you show it's false? Take your favorite example of a group $A$ with a nonnormal subgroup $B$; then $B$ is normal in $B$, but its image under the inclusion map $B\to A$ is not normal in $A$. On the contrary, if $G_2$ is abelian, then $\phi(N)$ would be normal in $G_2$ no matter whether $N$ is normal in $G$ or not.
Statement (a) is indeed true under the additional assumption that $\phi$ is surjective.
Your proof of (b) is correct.
For (c) your argument is wrong: there is no $\phi^{-1}$ map in general. The notation $\phi^{-1}(K)$ just denotes $\{x\in G_1:\phi(x)\in K\}$, but does not imply the inverse map exists. And you should prove that $\phi^{-1}(AB)=\phi^{-1}(A)\phi^{-1}(B)$, in case you just consider the inverse image. But you're using $\phi^{-1}(\phi(g))=g$ and this invalidates the proof.
Set $H=\phi^{-1}(K)$ for simplicity and consider $g\in G$; for $x\in H$, you have
$$
\phi(gxg^{-1})=\phi(g)\phi(x)\phi(g)^{-1}
$$
and this belongs to $K$, because $\phi(x)\in K$ by assumption and $K$ is normal. Therefore $gxg^{-1}\in H$.
