What is a Simple Group? I am working on this problem from class note:

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 $\phi (G) \subseteq N$.

Here are my two questions:
(1) Does the problem referring $G$ as a simple group? What is simple group? I did not see its definition in the text prior this problem. Note this is just a class note and not a commercial-grade textbook, therefore order getting messed up is understandable.
(2) I know that most likely I need to use Lagrange's $|G| = |G/H| \cdot|H|$ and therefore $|H|$ divides $|G|$, and also $G/\ker (\phi) \cong \phi (G)$ and therefore |$G/\ker (\phi)| = |\phi (G)|$, but after those I am lost. A gentle and friendly explanation would therefore be very much appreciated.
Thank you very much for your time and help.
EDIT: ~~~~~~~~~~~~~~~~~~~~~~ 
See my own answer after combining hints & helps from @Tim Raczkowski, @Blah and others. Thanks for all your time and effort.
 A: $$
G \rightarrow G' \rightarrow G'/N
$$
is a homomorphism to a group with 2 elements. It's kernel is either $G$ or $\{1\}$ since no other normal subgroups exist
Now you have to use $|G|\neq 2$.
A: From Blah's answer. Let $\sigma\,:G'\to G'/N$ be the canonical homomorphism, and $\psi=\sigma\circ\phi$. Since, $|G'/N|=2$ and $|G|>2$, cannot be injective. So $\ker\psi=G$. In other words, $\psi\left(G\right)=\{eN\}=N$. This implies $\sigma\left(\phi\left(G\right)\right)=N$, which means $\phi\left(G\right)\subseteq N$.
A: Having combined all the tips and helps from @Tim Raczkowski, @Blah and others, here is the wrap-up of what I made up so far:
(1) Let $\phi$ be homomorphism $\phi:G \to G′$ and $\sigma$ be the canonical homomorphis $\sigma:G′ \to G′/N$, and let the composition of two mappings be $\psi = \sigma \cdot \phi$:
$$G \overbrace{\underbrace{\to}_{\phi} G′\underbrace{\to}_{\sigma}}^{\psi} G′/N.$$
(2) Since $|G′/N|=2$ and $|G|\neq 2$, $\psi$ cannot be injective, meaning that $ker(\psi)\neq \{1\}$. 
(3) Since $G$ is simple whose normal subgroup is either $G$ itself or $\{1\}$, and since $ker(\psi)$ is normal subgroup of $G$, therefore it has to be $ker(\psi)=G$.
(4) In other words, $\psi(G)={eN}$, where $e$ is the neutral element of $G'$ and ${eN}$ is the neutral element of $G′/N$, that is, $\psi(G) = \sigma (\phi (G)) = {eN} = N$.
(5) By definition of $\sigma$:
$$\begin{align}
\sigma : G' &\to G'/N \tag{5a}\\
\sigma : \phi(G) &\to \phi(G)/N, \quad  \text{since} \ \phi(G) \subseteq (G) \tag{5b}\\
\sigma : \phi(g) &\mapsto \underbrace{\phi(g)N}_{= \ N}, \quad \forall g \in G \tag{5c}\\
\end{align}$$
(6) From $\phi(g)N = N$ we are forced to conclude that $\phi(g) \in N$, for otherwise $\phi(g)N \neq N$. 
(7) Since $\forall \phi(g) \in \phi(G) \rightarrow \phi(g) \in N$ therefore $\phi(G) \subseteq N. \blacksquare$
