I am reading Kassel's Quantum Groups book (the chapter on Drinfeld doubles). In it, there is the following claim:
If $H,K\subseteq G$ are groups such that $\forall g\in G$, $\exists!(y,z)\in H\times K$ such that $g = yz$, then we have maps $H\times K\rightarrow H$ (denoted $(y,z)\mapsto z\cdot y$) and $H\times K\rightarrow K$ (denoted $(y,z)\mapsto z^y$).
The claim is that from the associtivity in $G$, namely $(zz')y = z(z'y)$ and $z(yy') = (zy)y'$, I can derive relations like
$(zz')\cdot y = z\cdot (z'\cdot y)$ and $z\cdot 1 = 1$, $z^1 = 1$
or $(zz')^y = z^{z'\cdot y}{z'}^y$ and $z^{yy'}= (z^y)^{y'}$ etc. that make these into group actions of H and K on each other.
I am extremely annoyed, because I can prove some of these properties from each other, for instance
$\left[z\cdot (z'\cdot y)\right]{z'}^yz^{z'y}=(z\cdot z'y)z^{z'y}=z(z'y)=(zz')y = \left[(zz')\cdot y\right](zz')^y$ shows that $z\cdot (z'\cdot y) = (zz')\cdot y\iff (zz')^y = {z'}^yz^{z'y}$
but I can't seem to prove one of them just from the properties of $G$.
Surely this is easy but I am just not seeing it.
Never mind, I got it!
One needs to use the uniqueness property of the $(y,z)\in H\times K$ such that $yz = g$, and prove the relations in pairs!
For instance, we have $(zz')y = \left[(zz')\cdot y\right](zz')^y$ and also $z(z'y) = z(z'\cdot y){z'}^y = \left[z(z'\cdot y)\right]{z'}^y = \left[z\cdot (z'\cdot y) z^{z'\cdot y}\right]{z'}^y = \left[z\cdot (z'\cdot y)\right]z^{z'\cdot y}{z'}^y$
Now $((zz')\cdot y, (zz')^y)\in H\times K$ and $((z\cdot (z'\cdot y), z^{z'\cdot y}{z'}^y))\in H\times K$ are both pairs of elements that, when multiplied (in G) produce the same element $(zz')y = z(z'y)\in G$. Thus, they are equal in $H\times K$ by uniqueness.