Jordan Decomposition in a Linear Algebraic Group
Let \(G\) be a linear algebraic group and \(g\in G\). There are unique commuting elements \(g_s,g_u\in G\) such that \[ g=g_sg_u=g_ug_s, \] where \(g_s\) is semisimple and \(g_u\) is unipotent in any faithful linear representation of \(G\). The decomposition is functorial: for every homomorphism \(f:G\to H\), \[ f(g)_s=f(g_s),\qquad f(g)_u=f(g_u). \]
Proof
Embed \(G\) as a closed subgroup of \(\operatorname{GL}(V)\). In \(\operatorname{GL}(V)\), ordinary multiplicative Jordan decomposition gives unique commuting polynomial expressions \(g_s=P(g)\) and \(g_u=Q(g)\) with \(g_s\) semisimple and \(g_u\) unipotent. The polynomials can be chosen from the finite subalgebra generated by \(g\) and are intrinsic to \(g\).
Because \(G\) is Zariski closed and stable under the algebraic operations that define these polynomial expressions, \(g_s\) and \(g_u\) belong to \(G\). Uniqueness in \(\operatorname{GL}(V)\) implies uniqueness in \(G\). If \(f:G\to H\) is a homomorphism, then \(f(g_s)\) and \(f(g_u)\) are commuting semisimple and unipotent elements whose product is \(f(g)\), so uniqueness gives the functorial identities.