TITLE: Surprisingly short or elegant proofs using Lie theory
QUESTION [71 upvotes]: Today, I was listening to someone give an exhausting proof of the fundamental theorem of algebra when I recalled that there was a short proof using Lie theory:

A finite extension $K$ of $\mathbb{C}$ forms a finite-dimensional vector space over $\mathbb{C}$, so the group of units $K^\times$ would be of the form $\mathbb{C}^n\setminus\{0\}$, which is simply connected for $n>1$. Since the operation on $K^\times$ is essentially just multiplication of polynomials over $\mathbb{C}$, it must be a Lie group. In sum, if $n>1$, then $K^\times$ is a simply connected abelian Lie group, thus isomorphic (as a Lie group) to $\mathbb{C}^n$, which is absurd (since $\mathbb{C}^n$ is torsion-free). Thus, $n=1$.

What other examples are there of theorems which yield such short or elegant proofs by appealing to Lie theory?
To clarify the criteria: I'm looking for (nontrivial) theorems that are usually stated in terms outside of Lie theory (e.g. the fundamental theorem of algebra) that can be proven in a particularly short or elegant way using Lie groups or Lie algebras.

REPLY [4 votes]: I originally intended to write this as a comment but it turned out to be too long. The following is a proof of the fundamental theorem of algebra similar to OP's but which does not rely on the correspondence between connected, simply-connected Lie groups and Lie algebras but on somewhat easier results on Lie theory and some topological facts.
Suppose $K$ is a non-trivial field extension of $\Bbb{R}$ of degree $n$. Then, $K^\times$ is a connected, abelian real Lie group of dimension $n$. Commutativity implies that the exponential map $\mathrm{exp}:\Bbb{R}^n\to K^\times$ is a group morphism, and by connectedness we get that it is surjective and thus a covering map. The kernel $\Gamma$ of the exponential map is then a discrete subgroup of $\Bbb{R}^n$, isomorphic to $\Bbb{Z}^d$, and so $K^\times \simeq \Bbb{R}^{n-d}\times (S^1)^d$.
We know that $d\geq 1$ since there is torsion in $K^\times$, and in fact we must have $d=1$, for otherwise there would be $n^d$ $n$-th roots of unity in $K$. Therefore $K^\times \simeq \Bbb{R}^{n-1}\times S^1$ but also $K^\times \simeq \Bbb{R}^n\setminus\{0\}$, and we get that $n=2$ for otherwise $K^\times$ would be simply-connected.