# Under a canonical map, do normal subgroups map to normal subgroups?

Let $G$ be a group, $K\unlhd G$, $N\leq K$, and $N\unlhd G.$ Show that $K/N\unlhd G/N$.

My work is as follows. Define the canonical map $\pi:G\to G/N$. Then since $N\leq K\leq G$, we have $K\mapsto K/N$.

If $\pi$ preserves normal subgroups under mapping, then we are done. However, I can't find this statement in Artin, so I'm not sure if it's valid.

Is the statement below true, and if so why is it true?

If $\pi:G\to H$ is a canonical map and $N\unlhd G$, then $\pi(N)\unlhd H$.

Your last statement is in fact true for any surjective map. To see this, let $n\in N$ and $h\in H$; since $\pi$ is surjective, $h=\pi(g)$ for some $g\in G$, and thus $$h\pi(n)h^{-1}=\pi(g)\pi(n)\pi(g)^{-1}=\pi(gng^{-1})\in \pi(N)$$since $N$ is normal in $G$. This proves that $h\pi(N)h^{-1}\subset\pi(N)$ for all $h\in H$, and thus that $\pi(N)$ is normal in $H$.