Let $G$ be a nontrivial group. Show that $G$ is simple if and only if, for every group $H$ and homomorphism $f:G→H$, either $f$ is trivial or $f$ is injective.
So I have already proved that if $f$ is trivial, then given any normal subgroup $G'$ of $G$, $G'$={1}. But I have problem proving that if $f$ is injective, then given any normal subgroup $G'$ of G, $G'=G$. Any hint would be appreciated!
Best Answer
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G \to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=\ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.