I've read books which have this statement, without explanation : 'every finite group is an algebraic group'. I'm trying to understand what exactly they mean. The definition I have in my mind of a group scheme $ G $ over a base field $ k $ is a representable functor $$ G : (\text{Sch}/k)^{op} \rightarrow (\text{Grp}) $$ I discussed this with my professor and here's what he explained to me (which I find very beautiful):


Suppose $ G $ is a finite group. Consider the affine scheme $ X\_G = \coprod\_{g \in G} \text{spec} (k) = \text{spec} ( \prod\_{g \in G} k) $ which is the disjoint union of $ |G| $ copies of $ \text{spec} (k) $. This gives the corresponding Yoneda functor and for any $ k $-scheme $ S $, we have $$ X\_G(S) = \text{Hom} (S, X\_G) = \text{Hom}\_{k} \left( \prod\_{g \in G} k, \Gamma(S, \mathcal{O}\_S) \right) $$ Now, a $ k $-algebra morphism from $ \prod\_{g \in G} k $ to $ \Gamma(S, \mathcal{O}\_S) $ is determined by knowledge of the images of the idempotents $ e\_g \in \prod\_{g \in G} k $. So $$ X\_G(S) = \{ (s\_g)\_{g \in G} | s\_g \in \Gamma(S, \mathcal{O}\_S) , \sum s\_g = 1, s\_g s\_{g'} = 0, (s\_g)^2 = s\_g \} $$ The group structure is given by $ (s\_g) \cdot (t\_g) = (u\_g) $ where $ u\_g = \sum\_{hk=g} s\_h t\_k $. The fact that $ (u\_g) $ also satisfies the property of being in $ X\_G(S) $ is clear from computation. $ X\_G(S) $ indeed becomes a group with identity given by $ (a\_g) : a\_e = 1 $ and $ a\_g = 0 $ when $ g \neq e $ and the inverse of $ (s\_g) $ is $ (t\_g) $ with $ t\_g = s\_{g^{-1}} $.


So my (poorly phrased) question: Is this the only way to view a finite group as a group scheme or are there other ways too?


For context, this really came up in a discussion about quotients in algebraic groups. If $ G $ is $ GL\_n $ and $ T $ is a maximal torus with normalizer $ N\_G(T) $, then $ N\_G(T)/T = S\_n $, the symmetric group on $ n $ letters.



Viewing a finite group as a group scheme
I don't think that there is a different natural construction of a group scheme associated to a finite group. You can assume that authors mean this construction unless otherwise stated.


Notice, however, that this construction can be also done in the functorial setup, and a little bit more general and elegant as well:


Let $S$ be any base scheme. If $G$ is any set, define the functor $G(-) : (\mathsf{Sch}/S)^{\mathrm{op}} \to \mathsf{Set}$ by
$$G(X) := \{f : X \to G \text{ locally constant}\}.$$
The action on morphisms is clear. The functor $G(-)$ is representable by the $S$-scheme $\coprod\_{s \in G} S$, since a locally constant map $X \to G$ corresponds to a partition $X = \coprod\_{g \in G} X\_g$ into disjoint open subschemes, which corresponds to an $S$-morphism $X \to \coprod\_{g \in G} S$ (since $\mathsf{Sch}$ is [extensive](https://ncatlab.org/nlab/show/extensive+category)).


If $G$ is a group, then the functor $G(-)$ factors over $\mathsf{Grp}$. It follows that $\coprod\_{g \in G} S$ carries the structure of a group scheme over $S$. (There is no need to write down the multiplication map etc. We simply get it from the functorial characterization of group schemes.) If $G$ is finite and $S$ is affine, then $\coprod\_{g \in G} S$ is affine and of finite type over $S$.


