# Lie-Kolchin theorem, unipotent matrices and Ad-unipotent elements

Let $$G\le \text{SL}(n,\mathbb R)$$ be a connected real matrix Lie group, not necessarily solvable

The two basic questions I have, and I have been told by people that they "follow from" Lie-Kolchin theorem (or some other theorems from Lie theory) are

1.1 How to show $$u\in G$$ is a unipotent matrix (all eigenvalues are equal to $$1$$) $$\iff Ad_u:Lie(G) \to Lie(G)$$ is a unipotent map (all eigenvalues are equal to $$1$$)

or more generally

1.2 How to show $$u\in G$$ is a unipotent matrix (all eigenvalues are equal to $$1$$) $$\implies$$ for any representation $$\rho:G\to \text{GL}(V)$$, $$\rho(u)$$ is a unipotent linear transformation on $$V$$.

The second question I have is

2.Let $$H\lhd G$$. How to see that for any quotient map $$f:G\to G/H$$ sends Ad-unipotent elements to Ad-unipotent elements.

One obstruction for my questions is the conditions of Lie-Kolchin theorem. At least the versions I have seen are only for solvable groups over algebraically closed fields. I want the questions above to be answered at least for groups like $$G=\text{SL}(n,\mathbb R)$$.

Even if the conditions of Lie-Kolchin can be "smartly manipulated", I still don't see how to relate Lie-Kolchin theorem to the questions above. Since Lie-Kolchin theorem only talks about simutaneous eigenvector for all elements of a group and I don't see how unipotency comes into play. There is also a chance that I heard the term "Lie-Kolchin theorem" wrong. So please feel free to use any other parts of Lie theory if applicable. Please also feel free to answer any one of them if you are not sure about the others.

If $$G$$ is an algebraic group, then this seems to be a standard fact, as in Theorem 2.4.8 of T.A.Springer's linear algebraic group book.