Properties of Algebraic Groups Let $G(X_1, \cdots, X_n) \subseteq GL(V)$ denote the smallest algebraic group containing $\{X_i\}_{i=1}^{n} \subseteq GL(V)$, where $V$ is a finite-dimensional vector space over $\mathbb{C}$.
Let $S$ be a semisimple matrix, and $U$ a unipotent matrix (over $V$). It is known that $G(S)$ is a quasitorus (isomorphic to a product of $(K^{\times})^{n}$ with a finite group), and in particular a commutative group. Similarly, $G(U)$ is isomorphic to $(K^{m}, +)$, and in particular also commutative.
Suppose $S,U$ commute. 


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*Why is $G(S,U)$ commutative? 

*Why does it equal $G(S)G(U)=\{su | s\in G(S), u \in G(U) \}$?
Let $R:G \to GL(U)$  be some linear representation of an algebraic group $G$. Using the notation of the previous questions, 


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*Why is it true that $R(G(A))=G(R(A))$ for any $A\in G$?


Those questions are based on intuitive statements found in "Lie Groups and Algebraic Groups" by Onishchik and Vinberg. I am not sure how to prove them formally.
 A: The key to my questions was the wonderful book "Linear Algebraic Groups" by Armand Borel (specifically, page 57).
First, a lemma (1 of 2): if $M$ is a (not necessarily algebraic) subgroup of an algebraic $G$, then $G(M) = \overline{M}$ (the closure is in the Zariski topology). Proof of lemma 1: $G(M)$ is closed so it contains $\overline{M}$. $\overline{M}$ is a group: it contains the identity since $M$ does. It contains the inverse of each element since $x \to x^{-1}$ is a homeomorphism on the big group $G$ which shows $\overline{H}^{-1} = \overline{H^{-1}} = \overline{H}$. And it is closed under products: $x \in H \implies x\overline{H} = \overline{xH} = \overline{H} \implies H\overline{H} = \overline{H}$. We need to show $\overline{H}\overline{H} \subseteq\overline{H}$, so take $y\in \overline{H}$. We have $Hy\subseteq \overline{H} \implies \overline{H}y=\overline{Hy}\subseteq \overline{H} \implies \overline{H}\overline{H} \subseteq \overline{H}$. $\blacksquare$


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*Questions 1+2:


First I'll show that $G(U), G(S)$ commute. The abstract groups generated by $U$ and $S$, $\langle U \rangle = \{ U^k | k \in \mathbb{Z} \}$,$\langle S \rangle = \{ S^k | k \in \mathbb{Z} \}$ commute since $U,S$ commute (I mean that the commutator group $(\langle U \rangle, \langle S \rangle)$ is trivial). It will follow from the following lemma, applied to $M=\langle U \rangle, N= \langle S \rangle$, that $G(S),G(U)$ commute:
Lemma 2: If $M,N\subseteq G$ are subgroups of $G$, then the commutator subgroup $(M,N),(\overline{M},\overline{N})$ have the same closure. (In our case, $(M,N)={e}$, which is closed already.) Proof of lemma: Consider the algebraic homomorphism $c:G\times G \to G$ defined as $c(x,y)=xyx^{-1}y^{-1}$. $M \times N$ is dense in $\overline{M} \times \overline{N}$, so $c(M \times N)$ is dense in $c(\overline{M} \times \overline{N})$, which in turns shows $G(c(M \times N)) = G(c(\overline{M} \times \overline{N}))$. But by lemma 1, those 2 groups are the closures of $c(M \times N), c(\overline{M}, \overline{N})$. $\blacksquare$
The fact that $G(S)$ and $G(U)$ commute shows that the map $(s,u) \to (su)$ from $G(S) \times G(U) \to G$ is indeed a homomorphism, and the image of a homomorphism is an algebraic group too (well-known fact), so $G(S)G(U)$ is a subgroup.
So $G(S)G(U)$ is a commutative algebraic group, and it contains $S,U$, and it is evident that it is the minimal group containing both $S,U$. So $G(S,U)=G(S)G(U)$ and we're done.


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*Question 3:


One direction is trivial: $R(G(A))$ contains $R(A)$ and is an algebraic group, so it must contain the smallest algebraic group containing $R(A)$, which is $G(R(A))$ by definition.
The second inclusion is as follows: $R^{-1}(R(G(A))$ is closed (as the preimage of a closed set by a continuous function), and it contains $A$, so it must contain $G(A)$ - the smallest algebraic subgroup containing $A$, and so: $R(G(A)) \subseteq R(R^{-1}(R(G(A)))) \subseteq R(G(A))$.
Note: the more general equality $R(G(M))=G(R(M))$ when $M$ is any subset of $G$, with the same proof.
