Show $\phi:H\to HN/N$ is one-to-one.

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Can anyone help me with the question? I am stuck on showing this is one-to-one.

For subgroup $H$ of $G$ and normal subgroup $N$ in $G$, show the function $\phi:H\to HN/N$ defined by $\phi(h)=hN$ is one-to-one.

My work:

Let $\phi(h_{1})=\phi(h_{2})$. Then $h_{1}N=h_{2}N$. Then there is an $n\in N$ such that $h_{1}=h_{2}n$.

I feel like I need to use the fact that $h_{2}\in HN$?