Problem from "The Theory of Finite Groups" by Kurzweil and Stellmacher

Question

I am currently working my way through The Theory of Finite Groups by Kurzweil and Stellmacher and came across the following question in the first chapter:

Let $G$ be simple, $|G| \ne 2$, and $f : G \to H$ be a homomorphism. If $A \unlhd H$ is a normal subgroup of index $2$, then $\text{range}(f) \le A$.

Please could someone help me to prove this statement? It doesn't seem like it should be too difficult but having spent some time on it I'm still getting nowhere.

Thanks!

Answer 1

Look at the composition $$ G\xrightarrow{f} H\xrightarrow{\pi}H/A $$ where $\pi$ is the canonical projection. Then $K:=\ker(\pi\circ f)\unlhd G$ is a normal subgroup as the kernel of a homomorphism, and since $G$ is simple, there are two cases, either $K=G$, or $K=\{e\}$.

Investigate these two cases, mouse over below if you get stuck.

If $K=G$, $\operatorname{range}(f)\subseteq\ker(\pi)= A$. If $K=\{e\}$, the composition is injective, so gives an embedding of $G\hookrightarrow H/A$. So $|G|\leq 2$ since $|H/A|=2$, and by assumption $|G|=1$, necessarily, so $G=\{e\}$, and the claim is trivial.

Answer 2

If $G$ is abelian simple group, it should be cyclic of prime order. Hence $|G|$ is odd prime, so $f(G)$ would be of order $p$ or would be identity. The first case with $|H/A|=2$ forces that $f(G)$ must be inside $A$.

Assume $G$ is non-abelian simple group. Then hypothesis implies two things:

$$[G,G]=G \mbox{ and } [H,H]\subseteq A $$ Apply homomorphism to reach at your conclusion.