I am trying to understand the following proof. Claim: The subgroup $N < G$ is normal iff the relation $\equiv$ (N) respects products and inverses.
Proof: Suppose $N$ is normal, and suppose that $g \equiv g'$ (N) and that $h \equiv h'$ (N).
Then $(gh)^{-1}(g'h') = h^{-1}(g^{-1} g')h' = [h^{-1}(g^{-1}g')h] (h^{-1}h')] \in N.$
Also, $g \equiv_L g'$ (N) iff $g^{-1} \equiv_R (g')^{-1}$ (N) but if $N$ is normal then the two relations are the same. The converse is to be found in the textbook.
Two questions I have:
(1) Why start off with $(gh)^{-1}(g'h')$? While I understand how the two equalities hold, I don't understand the reasoning behind starting with $(gh)^{-1}(g'h')$.
(2) I don't understand how $[h^{-1}(g^{-1}g')h] (h^{-1}h')] \in N$.
If someone could please explain in detail I would be very thankful.
For (1), the intent is to show that $gh \equiv g^\prime h^\prime (N)$. This means that $(gh)^{-1}g^\prime h^\prime\in N$.
For (2), we know that $h^{-1}h^\prime\in N$ because this is the definition of $h\equiv h^\prime (N)$. Likewise, $g^{-1}g^\prime\in N$. Then, since $N$ is normal, we have that $h^{-1}(g^{-1}g^\prime)h\in N$. Now the product of these two elements must also belong to $N$, just because $N$ is a subgroup.