Proof that, if factor groups are equal, then the subgroups are equal

I'm reading a proof of the Correspondence Theorem and I'm seeing a step for which I'm not sure of the justification. So suppose that $G$ is a subgroup and $K< H_{1},H_{2} <G$, and $K\trianglelefteq G$. The move the proof makes is to say, "Since $H_{1}/K = H_{2}/K$ then $H_{1}=H_{2}$."

I've tried proving this by the following,

Suppose $h_{1}\in H_{1}$ so that $h_{1}K \in H_{1}/K = H_{2}/K$ and so there is some $h_{2}$ such that $h_{1}K = h_{2}K$. Since $h_{1}\in h_{1}K$ this implies there is some $k$ such that $h_{1}=h_{2}k$ ...

And at this point I don't see a way forward.

I think the big problem for me is that I don't have a criteria for "being in $H_{2}$".

• $H$ is precisely the union of the cosets contained in $H/K$... Nov 8 '14 at 1:26

Got it based on whacka's comment. If I were going to keep with my very flat-footed proof style, then I would remark that $K<H_{2}$ and therefore $h_{1}=h_{2}k\in H_{2}$.
But the proof suggested by whacka is shorter and easier: $H_{1}$ is the union of all cosets of $H_{1}/K$ is the union of all cosets of $H_{2}/K$ is $H_{2}$.