Commutative Algebraic Groups Split into Semisimple and Unipotent Parts
Let \(G\) be a commutative linear algebraic group over an algebraically closed field. Let \(G_s\) be the set of semisimple elements of \(G\), and let \(G_u\) be the set of unipotent elements of \(G\). Then \(G_s\) and \(G_u\) are closed algebraic subgroups, and multiplication induces an isomorphism of algebraic groups \[ G_s \times G_u \xrightarrow{\sim} G. \]
The subgroup \(G_s\) is diagonalizable, and \(G_u\) is unipotent.
Proof
For every \(g\in G\), the Jordan decomposition gives \(g=g_sg_u=g_ug_s\). Since \(G\) is commutative, the semisimple and unipotent parts of products are products of the corresponding parts: \[ (gh)_s=g_sh_s,\qquad (gh)_u=g_uh_u. \] Thus \(G_s\) and \(G_u\) are subgroups and every element has a unique product decomposition with factors in them.
Closedness follows by embedding \(G\) into a diagonal-plus-unipotent normal form inside some \(\operatorname{GL}(V)\): semisimple and unipotent parts are obtained by polynomial functions on the finite algebra generated by the elements. The product map is bijective with inverse \(g\mapsto(g_s,g_u)\), and these maps are morphisms. The semisimple commutative subgroup is simultaneously diagonalizable, while \(G_u\) is unipotent by definition.