My question is about a point on page 18 of J.S.Milne's "Algebraic Groups- The theory of group schemes of finite type over a field." and specifically about the fact that the translation map induced on an algebraic group by a rational point is an isomorphism. To be more precise the statement is :
Let $(G,m)$ be an algebraic group over k. For each $a∈G(k)$, there is a translation map $l_a:G\stackrel{\sim}{\to}\{a\}\times G\to G$ where the second map is the group multiplication $m$ of the group scheme G.
For $a,b\in G(k)$ $l_a\circ l_b=l_{ab}$.
From this "composition law " argument and the identity element property of the distinguished rational point the fact that the translation is an isomorphism follows easily.
While this is very clear intuitively, I struggle to show it using the categorical definition of group schemes, i.e the commutativity of the diagrams representing associativity, identity element and inversion.
I would really appreciate any help !
$\DeclareMathOperator{\id}{id}$ Associativity is expressed as the commutativity of the diagram $$\require{AMScd} \begin{CD} G \times G \times G @>{m \times \id}>> G \times G\\ @V\id \times mVV @VVmV \\ G \times G @>{m}>> G \end{CD}$$ Now if you restrict to $\{a\}\times \{b\} \times G$, you get $$\require{AMScd} \begin{CD} \{a\} \times \{b\} \times G @>{m \times \id}>> \{ab\} \times G\\ @V\id \times mVV @VVmV \\ \{a\} \times G @>{m}>> G \end{CD}$$ Identifying $G \cong \{a\}\times\{b\}\times G$ as before, the "upper way" in this diagram is $m \circ (m \times \id) = l_{ab}$, whereas the "lower way" is $m \circ (\id \times m) = l_a \circ l_b$. By commutativity you get $$l_{ab} = l_a \circ l_b.$$