Cartan-Brauer-Hua Theorem | Abstract Algebra

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The Cartan-Brauer-Hua theorem

Posted: April 12, 2024 in Division Rings, Noncommutative Ring Theory Notes
Tags: Cartan-Brauer-Hua Theorem, Hua's identity, multiplicative group of a division ring, normal subgroup

For a division ring $D,$ we denote by $Z(D)$ and $D^{\times}$ the center and the multiplicative group of $D,$ respectively.

Let $D_1$ be a division ring, and suppose that $D_2$ is a proper subdivision ring of $D_1,$ i.e. $D_2$ is a subring of $D_1, \ D_2 \ne D_1,$ and $D_2$ itself is a division ring. Then $D_2^{\times}$ is clearly a proper subgroup of $D_1^{\times}.$ Now, one may ask: when exactly is $D_2^{\times}$ a normal subgroup of $D_1^{\times} ?$ The Cartan-Brauer-Hua theorem gives the answer: $D_2^{\times}$ is a normal subgroup of $D_1^{\times}$ if and only if $D_2 \subseteq Z(D_1).$ In particular, $D_2$ must be a field. One side of the theorem is trivial: if $D_2 \subseteq Z(D_1),$ then $D_2^{\times}$ is obviously normal in $D_1^{\times}.$ The other side of the theorem is not trivial, but it’s not hard to prove either. The proof is a quick result of the following simple yet strangely significant identity!

Hua’s Identity (Loo Keng Hua, 1949). Let $D$ be a division ring, and let $a,b \in D$ such that $ab \ne ba.$ Then

$a=(b^{-1}-(a-1)^{-1}b^{-1}(a-1))(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))^{-1}.$

Proof. First see that since $ab \ne ba,$ the four elements $a,a-1,b,$ and $a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1)$ are all nonzero hence invertible. Now,

$a(a^{-1}b^{-1}a-(a-1)^{-1}b^{-1}(a-1))=b^{-1}a-a(a-1)^{-1}b^{-1}(a-1)$

$=b^{-1}a-(a-1+1)(a-1)^{-1}b^{-1}(a-1)=b^{-1}a-b^{-1}(a-1)-(a-1)^{-1}b^{-1}(a-1)$

$=b^{-1}-(a-1)^{-1}b^{-1}(a-1). \ \Box$

Cartan-Brauer-Hua Theorem. Let $D_1$ be a division ring, and let $D_2$ be a proper subdivision ring of $D_1.$ If $D_2^{\times}$ is normal in $D_1^{\times},$ then $D_2 \subseteq Z(D_1).$

Proof. Suppose, to the contrary, that $D_2 \nsubseteq Z(D_1).$ Let $a \in D_1, b \in D_2$ such that $ab \ne ba.$ Then, since $D_2^{\times}$ is a normal subgroup of $D_1^{\times},$ all the elements

$b^{-1}, \ (a-1)^{-1}b^{-1}(a-1), \ a^{-1}b^{-1}a, \ (a-1)^{-1}b^{-1}(a-1)$

are in $D_2^{\times}$ and hence, by Hua’s identity, $a \in D_2.$ So every element of $D_1 \setminus D_2$ commutes with $b.$ Now let $c \in D_1 \setminus D_2.$ Then $ac \in D_1 \setminus D_2,$ because $a \in D_2,$ and therefore both $c,ac$ commute with $b.$ But then $a=acc^{-1}$ will also commute with $b$ and that’s a contradiction. $\ \Box$