Solving Simple Group Problem: Subset of Normal Subgroup of Index 2

[A.Magnus]

I am working on myself on a problem looks like this:

Let $G'$ be a group and let $\phi$ be a homomorphism from $G$ to $G'.$ Assume that $G$ is simple, that $|G| \neq 2$, and that $G'$ has a normal subgroup $N$ of index 2. Show that $\phi (G) \subseteq N$.

I have been asking around and here is what I got so far:
(1) Since $G'$ is simple group, $G'$ does not have any non-trivial normal subgroup.
(2) Since $ker(\phi)$ is a normal subgroup of $G$, therefore it is either $ker (\phi) = G$ or $ker (\phi) = \{1\}.$
(3) In the first case, $ker (\phi) = G$ means $\phi (G) = 1$, and therefore $\phi (G) \subseteq N$ and we are done.
(4) In the second case, $ker (\phi) = \{1\}$ means that $\phi$ is injective and therefore $\phi (G) \cong G.$ This reduces the problem into proving that $(G) \subseteq N$.
(5) ...

And after that I am totally stuck, would appreciate any friendly and line-by-line befitting a person still in learning stage. Thank you for your time and help. Happy holidays.

 

[PeroK]

Try thinking about what can you do with N and $\phi(G)$

 

[Dick]

Try thinking about what can you do with N and $\phi(G)$

Also think about $G'/N$.