Clearly the kernel of a group homomorphism is normal, but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.
This feels correct but isn't entirely obvious to me.
One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.
I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.
So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?
Best Answer
You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.