For simplicity we work over a fixed algebraically closed field $\mathbb{k}$ of char 0. Recall that a linear algebraic group $G$ is a closed subgroup of $GL(n,\mathbb{k})$, and we say $G$ is linearly reductive if $Rep(G)$ is semi-simple, i.e. all representations decompose into a direct sum of irreducible representations.
My question is: given two linealy reductive algebraic groups $G$ and $H$, how to show that $G\times H$ is also linearly reductive?
Let's consider finite dimensional rational representations only.
Best Answer
In general, an algebraic group $G$ is called reductive if its unipotent radical is trivial. In characteristic zero, this is equivalent to linearly reductive. It is easy to see that linearly reductive implies reductive, but the other direction is more difficult.
Remark. In a reductive group the semi-simple elements form a dense set. In particular, every homomorphism of a reductive group to a unipotent group is trivial.
Now consider an exact sequence
$ 1 \to N \to G \to H \to 1$
of algebraic groups. Then
(1) If $G$ is reductive, then so are $N$ and $H$.
(2) If $N$ and $H$ are reductive, then so is $G$.
In particular, a product of reductive groups is reductive.
Proof of (1): (a) The unipotent radical is a characteristic subgroup. Hence the unipotent radical $R$ of $N$ is a normal unipotent subgroup of $G$. Thus $R$ is trivial and so $N$ is reductive.
(b) The inverse image of the unipotent radical of $H$ is a normal subgroup $R$ of $G$, hence $R$ is reductive by (a). But then the semi-simple elements are dense in $R$ and thus the homomorphism $R \to H$ is trivial. Hence $R$ is trivial and so $H$ is reductive.
Proof of (2): The image of the unipotent radical $R$ of $G$ in $H$ is a unipotent normal subgroup, hence trivial. Thus $R \subset N$ is a normal subgroup, and so $R$ is trivial.
The case of linearly reductive groups. In characteristic zero there is a rather short proof of the following statement:
If there is an exact sequence $1 \to K \to G \to H \to 1$ such that $K$ and $H$ are linearly reductive, then $G$ is linearly reductive.
Proof: We use the following characterization of linear reductivity:
For any surjective homomorphism of $G$-modules $M \to N$ the induced homomorphism $M^G \to N^G$ is also surjective.
Now let $M \to N$ be a surjective homomorphism of $G$-modules. Since $K$ is linearly reductive the induced homomorphism $M^K \to N^K$ is surjective.
But $M^K$ and $N^K$ are $G/K = H$-modules and the map $M^K \to N^K$ is a surjective homomorphism of $H$-modules. Since $H$ is linearly reductive, $(M^K)^H \to (N^K)^H$ is surjective, and the claim follows, because $(M^K)^H = M^G$.