If $H$ is a subgroup of $G$, then the homomorphism $i: HN/N \to p(H)$, defined by $i(xN) = p(x)$ is a topological isomorphism

Question

Let $G$ be a topological group, let $N$ be a normal closed subgroup of $G$ and let $p: G \rightarrow G/N$ be the canonical homomorphism. I need proof that if $H$ is a subgroup of $G$, then the homomorphism $i: HN/N \rightarrow p(H)$, defined by $i(xN) = p(x)$ is a topological isomorphism.

So far, I have only been able to prove that i is an isomorphism. I would like suggestions to prove continuity and that i is an open map. Below is my proof:

1. $i$ is clearly surjective
2. Let $x = hnN, y = h'n'N \in HN/N$ such that $$\begin{align} i(hnN) = i(h'n'N) = i(hN) = i(h'N) &\implies p(h) = p(h') \\ &\implies p(hh^{-1}) = 1 \\ &\implies hh^{-1}N = N \\ &\implies hN = h'N = hnN = h'n'N.\end{align}$$

Answer 1

I managed to resolve it after a while.

Consider the map q: HN $\rightarrow$ p(H) given by q(x) = xN.

This map is continuous because is a restriction to map p: G $\rightarrow$ G/N moreover, it's easily show that ker(p) = N, Im(P) = p(H) and p is open.

By first isomorphism theorem, there exists a unique continuous isomorphism i: HN/N $\rightarrow$ p(H) such that q = i o g ; g: HN $\rightarrow$ HN/N is the canonical homomorphism.

Therefore, i is a topological isomorphism because q is open.