# Does the Lie group $G_2$ contain any normal subgroups?

A nonabelian Lie group is called a simple Lie group if it contains no nontrivial connected normal subgroups. On the other hand, a group is called simple if it contains no nontrivial normal subgroups. So a simple Lie group may not be simple as a group. For example, $$SU(2)$$ is a simple Lie group, but it is not a simple group because $$\{\pm I\}$$ is a normal (actually, central) subgroup.

The $$14$$-dimensional compact Lie group $$G_2$$ is a simple Lie group. Is $$G_2$$ a simple group? That is:

1. Does $$G_2$$ contain a nontrivial normal subgroup (which is necessarily disconnected)?

I am aware that $$G_2$$ has trivial center, so any such normal subgroup is not central.

Actually, my real question (which may be easier to answer) is:

1. Does $$G_2$$ contain a normal subgroup isomorphic to $$\mathbb{Z}_2$$?

As $$G_2$$ has rank two, it has a subgroup isomorphic to $$S^1\times S^1$$ and therefore does have subgroups isomorphic to $$\mathbb{Z}_2$$, but I don't know if any of them are normal.

If there is a disconnected normal subgroup $$N$$ of $$G_2$$, then its connected component containing the identity (call it $$N_0$$) is a connected normal subgroup. Since $$G_2$$ is simple (as a Lie group), it follows that $$N_0$$ is the identity. For every $$n \in N$$, $$n N_0 = \{ n \}$$ is a connected component of $$N$$. Hence $$N$$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $$n$$, the map $$G \to N$$ sending $$g \mapsto gng^{-1}$$ is a connected subset of $$N$$ containing $$n$$, and therefore consists of only the point $$n$$).