At the very beginning of his book on Algebraic Groups, Milne defines an algebraic group as a group object in the category of finite type $k$-schemes, and an algebraic subgroup as a (locally closed) $k$-subscheme such that the inclusion map is a homomorphism of group schemes.
The text then asserts without proof the following claim:
Let $(G, m_G)$ be an algebraic group over $k$ and $H$ a nonempty $k$-subscheme of $G$. If $m_G|_{H \times H}$ and $\operatorname{inv}_G|_H$ factor through $H$, then $H$ is an algebraic subgroup with the induced operations.
This doesn't seem obvious from the definitions. In particular, I run into issues trying to show the group identity factors through $H$. Attempting to mimic the proof for abstract groups, I argue as follows:
First we must show $e : \operatorname{Spec} k \to G$ factors through $H$. Since $H$ is nonempty, it has some $L$-valued point, say $h \in H(L)$, for some finite extension $L/k$. Since the inversion and multiplication maps factor through $H$, we can find a candidate identity $e' = m(h, inv(h)) \in H(L)$.
I think one could show $e'$ is Galois-equivariant and cite faithfully flat descent to show $e'$ is defined over $k$, and then proceed from there. This feels like overkill.
Alternatively, one could prove the whole claim using the functor of points perspective. But the location and brevity of this claim in the text suggests I am missing something obvious. Is there a simple direct or arrow-theoretic proof that the identity factors through $H$?
Let $e \in G$ be the unit point.
Consider the map $f=m_H \circ (\mathrm{id}_H,\mathrm{inv}_H) : H \rightarrow H$. It follows from the hypotheses that $f$ composed with the inclusion $i: H \rightarrow G$, is equal to $g \circ i$, where $g=m_G \circ (\mathrm{id}_G,\mathrm{inv}_G)$ is the constant map $G \rightarrow G$ equal to $e$. Thus the set-theoretical image of $i \circ f$ is $e$, but $i \circ f$ has values in the underlying space $H_0$ of $H$, so $e \in H_0$.
As the residual field of $e$ in $G$ is $k$, the same is true for $e$ in $H$, and thus $e_G: \mathrm{Spec}\,k \rightarrow G$ factors through $H$: that is your unit section.