This is a very classical topic in algebraic group theory. If $k$ is a field and $G$ is an quasi-simple algebraic group defined over $k$ which is $k$-isotropic, then the group $G^+\subseteq G(k)$ generated by the unipotent elements in the $k$-parabolic subgroups is 'in most cases' simple modulo its center. (An algebraic group is quasi-simple if it has no proper connected normal subgroups defined over $k$, and $k$-isotropic if it has a $k$-split torus.)
The precise result, due to J. Tits, is as follows [J. Tits, Algebraic and abstract simple groups, Ann. Math. 80 (2), 1964]. If $G$ is a quasi-simple $k$-isotropic algebraic group defined over a field $k$ with more than 5 elements, then $G^+/Z(G^+)$ is simple.
This reduces the question to the quotient $G(k)/G^+$. The computation of this quotient is known as the Kneser-Tits problem. If $G$ is simply connected, then the quotient is in many cases trivial.
In the example $G=SL_n$, over any field $k$ with enough elements we have $G(k)=G^+$ (the typical unipotents being the conjugates of the upper triangular matrices with 1 on the diagonal) and indeed $SL_n(k)/Z(SL_n(k))$ is simple (and $SL_n$ is simply connected as an algebraic group).
On the other hand, the adjoint group $PGL_n$ is also quasi-simple (in the sense of algebraic groups) but not simply connected, with $PGL_n(k)^+=PSL_n(k)$. The quotient $PGL_n(k)/PSL_n(k)=G(k)/G^+$ is isomorphic to $k^*/(k^*)^n$.
If the group $G$ is not $k$-isotropic, then $G(k)$ may have many nontrivial normal subgroups. Consider for example the special orthogonal group $G=SO_n$ for the standard bilinear form. For $n\geq 5$, this group is in most cases quasi-simple. For $k=\mathbb R$, the compact Lie group $G(k)$ is simple modulo its center. But if $k$ is a non-archimedean real closed field (eg. the field of nonstandard reals $^*\mathbb R$), then the matrices infinitesimally close to 1 generate a normal subgroup in $G(k)$.
