Abstract Algebra Normal Subgroups

Question

I have a basic question related to abstract algebra, I read a proof in a book and partly below:

$K$ is a normal subgroup of $G$ and $H_1,H_2$ are two subgroups of $G$ containing $K$. Suppose $H_1/K = H_2/K$, then for any $h_1 \in H_1, h_1K \in H_2K$, so $h_1K = h_2K$ for some $h_2 \in H_2.$ Then $(h_2)^{-1}h_1\in K.$

I'm a little confused about the last sentence, how do we prove $(h_2)^{-1}h_1\in K$ rigorously?

Answer 1

This just comes down to basic coset manipulations. You have

$$h_1K = \{ h_1 k : k \in K\}.$$

You always have $h_1 \in h_1K$. Therefore, if $h_1K = h_2K$, this means that there exists a $k \in K$ such that $h_1 = h_2k$. Then $h_2^{-1}h_1 = k \in K$, as required.