See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Theorem 5, Corollary and Example (2). Example (2) answers your question when $K=\mathbb{C}$. A similar argument (using a $BN$-pair) proves the assertion over any algebraically closed field, because one can construct a $BN$-pair for any split group. 

This proves that the group $G(K)^+$ generated by the unipotent elements has no noncentral normal subgroups. When $K$ is algebraically closed, $G(K)^+=G(K)$.