Proving $\phi (N)$ is a normal subgroup of $\phi (G)$

Question

Let $\phi \colon G → K$ be a group homomorphism and let $N$ be a normal subgroup of $G$. Show that $\phi (N)$ is a normal subgroup of $\phi (G$).

Wondering what a good way to go about this is... I know a lot of different small things that can be derived from this information but not a way I can utilize this information to arrive at the intended conclusion.

Answer 1

Let $n' \in \phi(N),g'\in \phi(G)$, $n \in N\; \text{and} \;g\in G $ be such that $n'=\phi(n)$ and $g'=\phi(g)$. Then $$g'n'g'^{-1}=\phi(g)\phi(n)\phi(g)^{-1}=\phi(g)\phi(n)\phi(g^{-1})=\phi(gng^{-1})\in\phi(N)$$ since by Normality of $N$ we have $gng^{-1}\in N$. Therefore $\phi(N) \triangleleft \phi(G)$.

Answer 2

$\phi(N)$ is invariant under any inner automorphism $i_{\phi(g)}$ defined by an element of $\phi(G)$, since $i_{\phi(g)}(\phi(N))=\phi(g)\phi(N)\phi(g)^{-1}=\phi(gNg^{-1})=\phi(N)$, since $gNg^{-1}=N$ by normality of $N$.