# Finite Normal Subgroups Of Connected Groups Are Central

The previous series on algebraic groups is over. Actually, I barely got to the root system and root datum of a reductive group, but I found that the whole slew of material was getting too complex to organise on this blog, which I feel is better for more self-contained posts. Instead, I have begun to write a set of notes in real LaTeX describing the root datum of a reductive group and root systems in general, most of which I have completed, and I will release a draft soon.

In the mean time, I will continue posts on algebraic groups by reviewing some theorem and then illustrating it with an application such an an exercises from some textbook. Let's start by reviewing a simple one: let $k$ be an algebraically closed field. Then an irreducible finite set in the Zariski topology on $k^n$ (as a classical variety) necessarily has one element. (A topological space is irreducible if it cannot be written as the union of two proper closed subsets.)

We added the disclaimer "as a classical variety" since over an algebraically closed field $k$ it is sufficient to consider the closed points of $\mathrm{Spec}(k[x_i])$—otherwise the statement would be false.

Back to the theorem: a finite irreducible subset of $k^n$ has one element. This is just because each point is closed. Now, what can we deduce from this? Given a connected algebraic group $G$ over $k$, a finite normal subgroup is contained in $Z(G)$, the center of $G$. (A connected component in the case of algebraic groups is the same thing as an irreducible component.)

Here is a proof: Let $H$ be the finite normal subgroup. For each $y\in H$, the morphism $G\to G$ given by $x\mapsto xyx^{-1}$ maps $G$ into $H$ (since $H$ is normal) and since $G$ is connected, the image of $G$ in $H$ is connected, so the image is a single point because $H$ is finite.

But under this map the identity maps to $y$, so the point is $y$ itself. Hence $xyx^{-1} = y$ for every $x\in G$, which is equivalent to saying that $y\in Z(G)$.