Equivariant fibre product Let $G$ be an algebraic group. Let $X$ and $Y$ be $S$-schemes such that $X$, $Y$ and $S$ are $G$-schemes and the structural morphisms are equivariant. My question is: Can the fiber product $P:=X\times_SY$ be given the structure of a $G$-scheme in a natural way, i.e. such that the projections $P\to X$ and $P\to Y$ are equivariant? I have searched through Mumford/Fogarty/Kirwan for a while now and could not find any such statement.
Is it false in general? Are there certain conditions under which it becomes true? I am usually only dealing with varieties over an algebraically closed field and my group is just the multiplicative group of the ground field, but I need the fiber product over a third variety in this case.
 A: This is true for abstract nonsense reasons. If $G$ is a group object in a category $\mathcal{C}$, then $G \times S$ is a group object in the slice category $\mathcal{C}_{/ S}$, and there is a natural bijection between $G$-actions on an object in $\mathcal{C}_{/ S}$ (considered as an object in $\mathcal{C}$) and $(G \times S)$-actions on the same object (considered as an object in $\mathcal{C}_{/ S}$), simply because there is a natural isomorphism $(G \times S) \times_S X \cong G \times X$. Thus, we may assume without loss of generality that $S$ is the terminal object in $\mathcal{C}$. But then $X \times Y$ has an obvious $G$-action if $X$ and $Y$ do: namely, the diagonal action.
More explicitly, suppose we have actions $G \times X \to X$, $G \times Y \to Y$, and $G \times S \to S$ and $G$-equivariant morphisms $X \to S$, $Y \to S$. Then, we can define a $G$-action $G \times (X \times_S Y) \to X \times_S Y$ as the unique morphism such that these equations hold:
\begin{align}
(G \times (X \times_S Y) \to X \times_S Y \to X) & = (G \times (X \times_S Y) \to G \times X \to X) \\
(G \times (X \times_S Y) \to X \times_S Y \to Y) & = (G \times (X \times_S Y) \to G \times Y \to Y)
\end{align}
