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)$.

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