Sorry for my bad English.
I'm having some trouble understanding the claim in Mumford's "Abelian Varieties" p.111.
Let $k$ be algebraically closed field, $X$ be group scheme over $k$, and $G\subset X$ be a finite subgroup scheme.
In this book, he says $X/G\times X/G\cong (X\times X)/(G\times G)$ as scheme.
Now we have natural map $(X\times X)/(G\times G)\to X/G\times X/G$ by universal property of the product or by $G\times G$-invariant of $X\times X\to X/G\times X/G$.
But I don't know how to construct $X/G\times X/G\to (X\times X)/(G\times G)$, so that I don't know how to prove $X/G\times X/G\cong (X\times X)/(G\times G)$.
Please tell me proof or hint, thanks.
To give a map $$X/G\times X/G\to X\times X/(G\times G)$$ we need to give for a scheme $S$ over $k$, the map $$X/G\times X/G(S)\to X\times X/(G\times G)(S)$$ i.e. $$X/G(S)\times X/G(S)\to X\times X/(G\times G)(S).$$
Let $a, b:S\to X/G$. There is an fppf-cover of $S'\to S$ of $S$ such that $a$ and $b$ lift to $a', b': S'\to X$.
The lifts $a', b': S'\to X$ are so that $S'\to S\xrightarrow{a} X/G$ agrees with $S'\xrightarrow{a'} X \to X/G$ and similarly for $b'.$
Now $a'\times b': S'\to X\times X$.
Consider $$a'\times b': S' \xrightarrow{a'\times b'} X\times X\to X\times X/(G\times G).$$
We observe by the fppf-descent that the morphism $a'\times b'$ descents to $S\to X\times X/(G\times G)$ which we denote by $a\times b$.
Observe that this is well defined i.e. does not depend on the lifts $a'$ and $b'$ of $a$ and $b$ respectvely.