# 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.

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.
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 $$gxg^{-1}\in H$$.