I'm stuck with the following problem. Can someone help me by providing a hint?
Suppose $G$ is simple and let $f$ be an homomorphism between $G \to H$. If $\#G\ne2$, $A\lhd H$, and $[H:A]=2$. Then $\text{Im}f$ is a subgroup of $A$.
I know that the kernel of the homomorphism is a normal subgroup of $G$ so either is the trivial subgroup or the entire group. The last case is trivial, because everything is send to $1$ in $H$, and clearly the trivial group is a subgroup of the $A$.The case when the kernel is trivial is my problem. Here we can conclude that the orden of the image is not two, because is the same as $G$ (is one to one) and also divides the orden of $H$, i.e., $\#H=[H:A]\#A=2\#A$ i.e., the image divides the even number. But from here I'm not sure of how proceed, any idea?
EDIT A is normal and so its inverse image is a subgroup and in this case in not too difficult to show that is also normal, since is en $G$ is either the trivial subgroup or the entire group. The idea is to show that it must to have the entire group. If not, $f^{-1}=\{1\}$ but from here i don't know how to get the contradiction. any idea of how to eliminate the possibility of $f^{-1}={1}$
Consider the composition $$ G\xrightarrow{f} H\xrightarrow{\pi}H/A. $$
Now $K:=\ker(\pi\circ f)\leq G$ is normal as the kernel of a group homomorphism, so either $K=G$, or $K=\{e\}$, since $G$ is simple. If $K=G$, then $f(G)\subseteq\ker\pi=A$, and you're done.
If $K=\{e\}$, then $\pi\circ f$ is an injection of $G\hookrightarrow H/A$. But $[H:A]=2$, so $|G|\leq 2$. By hypothesis, $|G|\neq 2$, so $|G|=1$. Then $G$ is trivial, and it follows $f(G)\subseteq A$.