Division structure of a torsor I was looking at torsors from nlab, and I got stuck at the trivialization part. I could prove the fact that the set underlying a $G$-torsor $\bar T$ is isomorphic to the set underlying $G$, the so called trivialization of $\bar T$. But then remark 2.13 stumped me. Before I proceed, some context of the notations, definitions and results.
A $G$-torsor $\bar T$ is a set $T$ and a group action $a:G\times T\to T$ such that the map $a\times \pi_2:G\times T\to T\times T$ is an isomorphism (where $\pi_2:G\times T\to T$ is the natural projection on the second coordinate). Then a theorem says that given $T$ is nonempty, then the set underlying $G$ is isomorphic to $T$, and such an isomorphism is called a trivialization. Let $\rho:G\to T$ be a trivialization.
Now onto the problems. They define a 'division' among the elements of $T$ by the map $$d:=\pi_{1} \circ \left( \rho^{-1} \times id \right): T \times T \rightarrow G \times T \rightarrow G$$ which seems kind of fine, yet iffy to me for the following reason. It seems fine because we can assume that the trivialization $\rho$ comes from the choice of some element $t\in T$, we can identity $\rho$ with $t$ itself, and call the map $t$ (sloppy language for sure, but helps me visualize what's going on, maybe). Then, $d(t_1,\ t_2)=\pi_1\circ (t^{-1}\times id)(t_1,\ t_2)=\pi_1(t^{-1}(t_1),t_2)=t^{-1}(t_1)$ which looks like division for sure, but we have forgotten the element $t_2$ altogether. Something fishy is going on, or I am being dumb.
So question 1 : Am I thinking correctly? If yes, where does the element $t_2$ vanish? If not, then what is the correct interpretation of said map?
Moving on, they define $$D:=\rho\circ d:T\times T\to T$$ and call this a 'division structure on $T$'.
Question 2: What is a division structure on a set? Maybe the answer lies in the answer of question 1, but I am really lost.
Finally, the map $D$ looks very odd. Suppose $(t_1,\ t_2)\in T\times T$, then $$D(t_1,\ t_2)=\rho\circ d(t_1,\ t_2)=\rho(\rho^{-1}(t_1))=t_1$$ and thus whatever value I take in the second coordinate, it vanishes into thin air.
Question 3: How does $D$ make sense?
Condensing all the questions into one question, what I really want to ask is : Am I misinterpreting the maps $d$ and $D$, or am I missing something?
Sorry for the long question, but I am really stuck and appreciate any help.
 A: If you've never worked with torsors before it's worth carrying around some examples in your head as you work through all these definitions. Here are two nice ones:

*

*the set of total orders on a finite set $X$ is a torsor for the symmetric group $\text{Aut}(X)$, and

*the set of bases of a finite-dimensional vector space $V$ is a torsor for the general linear group $GL(V)$.

A more straightforward but less categorical definition of a $G$-torsor is that it's a $G$-set $T$ on which $G$ acts freely and transitively; equivalently, such that for each $x, y \in T$ there is a unique $g \in G$ such that $gx = y$. (Take a minute to see why this holds in the two examples I just gave.) This $g$ is the "division" operation.
The definition of division given in the nLab cannot possibly be correct, because it makes no reference to either the action map or the group operation; you are correct that something fishy is going on here. If $\rho : G \to T$ is a trivialization, the condition $gx = y$ becomes $g \rho(h) = \rho(k)$, or equivalently $\rho(gh) = \rho(k)$, so $gh = k$, so $g = h^{-1} k$. So in terms of a trivialization, division is given by
$$d : (x, y) \mapsto \rho^{-1}(x)^{-1} \rho^{-1}(y)$$
which means it can be written as the composite
$$T \times T \xrightarrow{\rho^{-1} \times \rho^{-1}} G \times G \xrightarrow{h^{-1} k} G.$$
A "division structure" is just a map that axiomatizes the properties of the division operation $(h, k) \mapsto h^{-1} k$ on a group $G$.
For context, it's probably worth knowing that the reason the nLab is describing what should be a simple concept so tortuously is that torsors become much more interesting in categories other than $\text{Set}$, and in particular they no longer admit trivializations in general. For example we get nontrivial principal bundles. The set-of-bases example above generalizes to a construction called taking the frame bundle of a vector bundle.
