Why are unipotent elements of a linear Lie/algebraic group well defined?

Question

For a Lie group $G$ embedded in some $SL_n$, unipotent elements are defined to be those which satisfy $(x-I)^n = 0$ for some large enough $n$. However, this definition seems to depend on the manner in which $G$ is embedded inside $SL_n$. Is it true that for any two embeddings of $G$ in $SL_n$ and $SL_m$, the unipotent elements from the first representation are the same as the ones from the second representation? If not, can we strengthen assumptions on $G$ to make this happen, like require $G$ to be semisimple? What about when we additionally require $G$ to algebraic, and the corresponding embeddings to be algebraic as well?

Answer 1

To get this off the unanswered list.

If we're talking about algebraic groups, which you seem to indicate is at least one of your interests then yes, this is true. Note that unipotence in $\text{SL}_n$ is the same as that in $\text{GL}_n$. Thus, one can apply [Mil, Theorem 9.18].

[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.