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See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5 (due to Tits). We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also that $G$ is quasi-split, which is the case when $K$ is an algebraically closed field or a finite field. We construct a $BN$-pair taking for $B$ a Borel subgroup. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, Corollary of Theorem 5 says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no nontrivial noncentral normal subgroups. When $G$ is quasi-split and simply connected, $G(K)^+=G(K)$ (Steinberg). Thus $G(K)$ has no noncentral normal subgroups, which seems to answer the question over an algebraically closed field and over a finite field.

See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5 (due to Tits). We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also that $G$ is quasi-split, which is the case when $K$ is an algebraically closed field or a finite field. We construct a $BN$-pair taking for $B$ a Borel subgroup. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, Corollary of Theorem 5 says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no nontrivial noncentral normal subgroups. When $G$ is quasi-split and simply connected, $G(K)^+=G(K)$ (Steinberg). Thus $G(K)$ has no nontrivial noncentral normal subgroups, which seems to answer the question over an algebraically closed field and over a finite field.

See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5. We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We construct a $BN$-pair. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, the corollary says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no noncentral normal subgroups. When $K$ is algebraically closed, $K$ is infinite and $G(K)^+=G(K)$. Thus $G(K)$ has no noncentral normal subgroups, hence it has no nontrivial infinite normal subgroups. Thus $G$ has no nontrivial algebraic subgroups of positive dimension.

See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5 (due to Tits). We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also that $G$ is quasi-split, which is the case when $K$ is an algebraically closed field or a finite field. We construct a $BN$-pair taking for $B$ a Borel subgroup. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, Corollary of Theorem 5 says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no nontrivial noncentral normal subgroups. When $G$ is quasi-split and simply connected, $G(K)^+=G(K)$ (Steinberg). Thus $G(K)$ has no noncentral normal subgroups, which seems to answer the question over an algebraically closed field and over a finite field.

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, $G(K)^+=G(K)$.

See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5. We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We construct a $BN$-pair. Since the root system is irreducible, the Coxeter system is irreducible. If we denote by $G(K)^+$ the (normal) subgroup of $G(K)$ generated by the unipotent elements, the corollary says that $G(K)^+/(G(K)^+\cap Z(K))$ is either a nonabelian simple group or reduces to one element. This proves that then the group $G(K)^+$ has no noncentral normal subgroups. When $K$ is algebraically closed, $K$ is infinite and $G(K)^+=G(K)$. Thus $G(K)$ has no noncentral normal subgroups, hence it has no nontrivial infinite normal subgroups. Thus $G$ has no nontrivial algebraic subgroups of positive dimension.