Jordan decomposition of elements in non-connected component of algebraic group Let $G$ be a Linear Algebraic Group (over algebraically closed field). We know that the connected component $G^o$ is a normal subgroup of finite index in $G$. Let $g$ be an element of $G$ which is not in $G^o$.
I want to understand Jordan decomposition of $g$ as a product of semisimple and unipotent elements. 
Modification from earlier question:
Can I choose representatives for $G/G^o$ consisting of semisimple elements alone? Of course the answer is 'no' in char p. But how bad is the scene.
Thanks a lot.
 A: The question itself falls somewhat short of being "research-level", but maybe it's useful to expand a little on the comment by anon (which the proposed "answer" by Anupam doesn't improve on).    
The Jordan-Chevalley decomposition in an arbitrary linear (= affine) algebraic group $G$ over an algebraically closed field $K$ involves the fundamental insight that being defined by finitely many polynomial conditions somehow guarantees that $G$ contains the semisimple and unipotent parts of its elements while these are independent of the linear realization of $G$.      More precisely, any morphism $G \rightarrow H$ of algebraic groups takes semisimple (resp. unipotent) elements to semisimple (resp. unipotent) elements.  These ideas go back to a 1948 paper by Ellis Kolchin, but were only explored systematically by Chevalley when he laid foundations in the 1950s for the theory in arbitrary characteristic.    The older language of algebraic geometry can be updated to scheme language, but without serious effect on the Jordan decomposition idea.
As pointed out in the comment by anon, the canonical morphism of algebraic groups $G \rightarrow G/G^\circ$ has as image a finite group.
The latter can be viewed as a linear algebraic group, and its Jordan-Chevalley decomposition reduces to the usual elementwise decomposition in a finite group.   If the characteristic of $K$ is $p>0$, one gets an  element of order prime to $p$ times an element of order a power of $p$ (each being a power of the given element).    If char $K = 0$, on the other hand, all elements of the finite group are semisimple: this follows from linear algebra, since the minimal polynomial of a matrix of finite order divides some $x^n-1$ by Cayley-Hamilton and thus has distinct roots.   In this case, it's clear that a set of coset representatives for $G/G^\circ$ can alwqays be chosen to consist of semisimple elements: all unipotent elements of $G$ lie in $G^\circ$, so the semisimple parts of an arbitrary set of coset of representatives also form such a set.   
A: Based on the discussion, I think a reasonable answer is that, look at $G/G^o$ which is a finite group. For finite group Jordan decomposition is simple looking at the p-powers of the element and coprime-to-p powers of the element. 
