I am trying to prove the following statement.
Suppose we have groups $G$ and $H$. Let $\phi : G \to H$ be an onto homomorphism. If $N$ is a normal subgroup of $G$ and $\phi(N)$ is a subgroup of $H$, then $\phi(N) \triangleleft H$.
My attempt : Consider elements $g\in G$ and $n \in N$. Since $N$ is a normal subgroup of $G$, we have that $gng^{-1} \in N$. Then $\phi(gng^{-1}) \in \phi(N)$. Since $\phi$ is an homomorphism, it follows that $\phi(gng^{-1}) = \phi(g)\phi(n)\phi(g^{-1}) =\phi(g)\phi(n)\phi(g)^{-1} \in \phi(N)$ where $\phi(g) \in H$ and $\phi(n) \in \phi(N)$. Therefore $\phi(N) \triangleleft H$ as desired.
Is this a valid proof for the given statement?
Any advice would be appreciated.