In Fraleigh, there is what appears to be a classic theorem on cosets. I'm confused about the proof for the converse.
Theorem. Let $H$ be a subgroup of a group $G$. Then left coset multiplication is well defined by the equation $$(aH)(bH) = (ab)H$$ if and only if $H$ is a normal subgroup of $G$.
Proof. ...We now turn to the converse: if $H$ is a normal subgroup, then left multiplication by representatives is well-defined. Suppose we wish to compute $(aH)(bH)$. Choosing $a \in aH$ and $b \in bH$, we obtain the coset $(ab)H$. Choosing different representatives $ah_1 \in aH$ and $bh_2 \in H$, we obtain the coset $ah_1 b h_2 H$. We must show these are the same coset. Now $h_1 b \in H b = bH$, so $h_1b = bh_3$ for some $h_3 \in H$. Thus $$(ah_1)(bh_2) = a(h_1 b) h_2 = a(bh_3)h_2 = (ab)(h_3h_2)$$ and $(ab)(h_3h_2) \in (ab)H.$ Therefore $ah_1 b h_2$ is in $(ab)H$.
This seems weird to me. Why do we name our second representatives $ah_1$ and $bh_2$, rather than $a', b' \in G$? For example, in this post Arturo produces a proof for this,
Proof. Suppose $H$ is normal. Then $gH=Hg$ for every $g\in G$. If $xH=x'H$ and $yH=y'H$, then $$xyH = x(yH) = x(Hy) = x(Hy') = (xH)y' = > (x'H)y' = x'(Hy') = x'(y'H) = x'y'H$$ so the operation is well defined.
There must be some motivation for using this seemingly awkward way of picking representatives.