Cartan-Brauer-Hua theorem

Statement
Let $$K < L$$ be fact about::division rings (i.e., skew fields), with $$K$$ properly contained in $$L$$. Suppose $$K^*$$ is a normal subgroup of $$L^*$$. Then $$K^*$$ is a central subgroup of $$L^*$$, viz $$K^*$$ is contained in the center of $$L^*$$. In particular, this implies that $$K^*$$ is abelian and hence $$K$$ is a field.

Related facts in group theory

 * Normal not implies central
 * Central implies normal
 * Totally disconnected and normal in connected implies central
 * Normal of order equal to least prime divisor of group order implies central

Related facts about division rings

 * Every finite division ring is a field (originally proved by Wedderburn)
 * Bruck-Kleinfeld theorem

Any element inside commutes with any element outside
For nonzero elements $$x,y \in L^*$$, we denote by $$[x,y]$$ the multiplicative commutator $$xyx^{-1}y^{-1}$$ and by $$\! c_x(y)$$ the element $$xyx^{-1}$$.

We denote by $$c_x$$ the map $$y \mapsto xyx^{-1}$$. Here, $$x \in L^*$$ but $$y$$ is allowed to be zero.

Given: $$g \in K^*$$ and $$a \in L \setminus K$$.

To prove: $$[g,a] = 1$$.

Proof: The key idea is to play off the additive and the multiplicative structure against each other, and use the fact that the map $$y \mapsto xyx^{-1}$$ is an automorphism of both the additive and the multiplicative structure.

The finishing touch
Now, if $$K$$ is a proper subset of $$L$$, we will show that $$K^*$$ is contained inside the center. We already know that every element of $$K^*$$ commutes with every element of $$L \setminus K$$. So it suffices to show that any two elements of $$K^*$$ commute.

Let $$g,h \in K^*$$. Then take any $$a \in L \setminus K$$. Then, $$a + h \in L \setminus K$$. Thus, $$g$$ commutes with both $$a+h$$ and $$a$$. Hence $$g$$ must commute with the difference, which is $$h$$.