Composition at points in the domain of a rational group action

Products here are integral: for affine integral coordinate rings $A,B$, write two alleged zero-divisor factors in $A{\otimes }_{k}B$ using finite linearly independent coefficients in $B$. Specializing at each closed point of $\mathrm{Spec}A$ makes one factor zero because $B$ is a domain. The two vanishing loci are closed and cover the irreducible $\mathrm{Spec}A$ by [F3], so one factor is zero. This proves $A{\otimes }_{k}B$ is a domain. Put $T=G\times G\times X$, $a(u,v,z)=\alpha (v,z)$, and $b(u,v,z)=\alpha (u,a(u,v,z))$. The simultaneous domain $N$ of $a$ and $b$ is open and contains $(g,h,x)$. The rational map $c(u,v,z)=\alpha (uv,z)$ equals $b$ by the action identity. The closure of the graph of $(a,b,c)$ is contained in the closed locus where its last two coordinates coincide. Over $N$ it is exactly the graph of $(a,b,b)$: the latter is closed by [F1], contains the dense generic graph there, and that graph is dense in it because $N$ is integral. Consequently $c$ has the regular representative $b$ on $N$.

The morphism $s:G\times X\to T$ given by $s(t,z)=(t{h}^{-1},h,z)$ is a section of $(u,v,z)\mapsto (uv,z)$. Its inverse image of $N$ is an open neighbourhood of $(gh,x)$. On that neighbourhood $b\circ s$ is a morphism. It agrees with $\alpha$ on the dense open where $\alpha$ itself is defined, since there $c\circ s=\alpha$; this dense open intersects the neighbourhood because $G\times X$ is integral. Thus it represents $\alpha$ and extends its domain to $(gh,x)$. Its value is $b(g,h,x)=g\cdot (h\cdot x)$, as claimed.