Component group of stabilizer group of a nilpotent element

Question

Let $G$ be a semisimple complex algebraic group, and $\mathfrak{g}$ be its Lie algebra. $G$ acts on $\mathfrak{g}$ by adjoint action. Let $x$ be an nilpotent element in $\mathfrak{g}$, and $G(x)\subset G$ be its stabilizer group. Let $C(x)$ be the component group of $G(x)$, i.e., $C(x)=G(x)/G(x)^0$, where $G(x)^0$ is the identity component of $G(x)$.

For an arbitrary nilpotent element in $\mathfrak{g}$, is the group $C(x)$ always finite Abelian? If not, what is a countexample?

Answer 1

No. Example 16 of Sommers - A generalization of the Bala–Carter theorem for nilpotent orbits shows that, if $x$ is a subregular nilpotent element in $\mathsf G_2$, then $\operatorname C(x)$ is $\mathrm S_3$.

See also Centralizers of nilpotent elements in semisimple Lie algebras and @FrancoisZiegler's answer to Stabilizers for nilpotent adjoint orbits of semisimple groups. I would say that this question is nearly a duplicate of the former, but it is more focussed.