In the book by 'Algebraic Groups' by Milne, a right $G$- torsor over $S_0$ is a scheme $S$ faithfully flat over $S_0$ together with an action $S\times_{S_0}G\to S$ of $G$ on $S$ such that the map
$$(s,g)\mapsto(s,gs): S\times_{S_0}G\to S\times_{S_0}S $$
is an isomorphism of $S_0$-schemes. then There is an exercise that says, let $G\to Q$ be a faithfully flat homomorphism of algebraic groups with kernel $N$. The action $G\times_{Q}N\to G$ of $N$ on $G$ induces an isomorphism $G\times_{Q}G \to G\times N$ and so $G$ is a torsor under $N$ over $Q$.
Why does $G\times N \simeq G\times_{Q}G$ say its $N$ torsor when it should be $G\times_{Q}N \simeq G\times_{Q}G$?
Thank you for the help.
The homomorphism $G\rightarrow Q$ and its kernel $N$ are both over a field $k$. We have $G\times_{Q}G\simeq G\times N\simeq G\times_{Q}N_{Q}$, which says that $G$ is a torsor under $N_{Q}:=N\times Q$ over $Q$ (unadorned products are over Spec$(k)$). By an abuse of language, we just say that $G$ is a torsor under $N$ over $Q$.