Let $G$ be simple group, $|G|$ is not 2, and $ ϕ $ a homomorphism from $G$ to $H$. If $H$ contains a normal subgroup A of index 2, then $ ϕ(G) ≤ A $ .
I have been trying the following: $A$ is normal and maximal in $H$ since its index is 2(basic fact + correspondence theorem). $G$ is simple so it is isomorphic to its image so its image is a simple subgroup of $H$. Now I have to show somehow that its image must be a subgroup of $A$ and I can't find any reason it has to. I would very appreciate if someone could help me with this.
Proof: Let $\pi : H \to H/A$ be the natural quotient map, and let $f = \pi \circ \phi: G \to H \to H/A$. The kernel of $f$ is the set of $g \in G$ such that $\pi(\phi(g))$ is the coset $1_HA = A$; that is the set of $g \in G$ such that $\phi(g) \in A$, which is exactly $\phi^{-1}(A)$. Since $G$ is simple, we have that $\phi^{-1}(A) = \ker(f)$ is either $\{1_G\}$ or $G$ (since kernels are normal subgroups). Since $|G| > 2$ and $|H/A|=[H:A] = 2$, the pigeon hole principle says that $f$ cannot be injective and so $\phi^{-1}(A)=\ker(f) > \{1_G\}$; hence $\phi^{-1}(A) = \ker(f) = G$. Thus $\phi(G) = \phi(\phi^{-1}(A)) \leqslant A$.