Cartan–Brauer–Hua theorem

Remarks

A short proof works with two conjugates. Take $a \in K$ and $b \in D \setminus K$; then $b \ne 0$ and $1 + b \ne 0$, so $k_1 = b a b^{-1}$ and $k_2 = (1+b) a (1+b)^{-1}$ lie in $K$. Expanding $(1+b)a = k_2(1+b)$ and using $ba = k_1 b$ gives $(k_1 - k_2) b = k_2 - a$, so $k_1 = k_2$ (otherwise $b \in K$), hence $a = k_2 = k_1$ and $ab = ba$. So every element of $K$ commutes with $D \setminus K$; if $K \ne D$, every element of $K$ is a difference of two elements of $D \setminus K$, so it also commutes with $K$, and $K \subseteq Z(D)$. Example: in the real quaternions $\mathbb{H}$, the subfield $\C = \R + \R i$ is neither $\mathbb{H}$ nor central, and indeed $(1+j)\, i \,(1+j)^{-1} = -k \notin \C$.

Going deeper

For a student

The theorem is a division-ring analogue of the fact that a normal subgroup is a union of conjugacy classes. It is often applied with $K$ the subring generated by a conjugation-invariant set, such as the center together with some commutators. Its proof is elementary, in contrast with deeper results such as Wedderburn's little theorem, and it has a companion in Hua's identity, another elementary tool for division rings.