Cartan-Brauer-Hua theorem

This page describes additional conditions under which a subgroup property implication can be reversed, viz a weaker subgroup property, namely Normal subgroup (?), can be made to imply a stronger subgroup property, namely central subgroup
View other subgroup property implication-reversing conditions

Statement

Let $K<L$ be 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

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

Proof

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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.

Step no.	Assertion/construction	Given data/assumptions used	Previous steps used	Explanation	Commentary
1	$g,a,a+1$ are all in $L^{*}$, so the notations $c_{g}$, $[g,a]$, and $[g,a+1]$ make sense.	$g\in K^{*},a\in L\setminus K$.		[SHOW MORE] $g\in K^{*}$, so $g\in L^{*}$. Further, $a\in L\setminus K$, so $a$ is nonzero too. Further, since $a\notin K$, we know that $a\neq -1$ (since $-1$ would be in any skew field) and hence $a+1\neq 0$.	The choice of $a$ and $a+1$ allows us to play on addition.
2	$\![g,a]\in K^{*}$ and $\![g,a+1]\in K^{*}$.	$K^{*}$ is normal in $L^{*}$	Step (1)	[SHOW MORE] The commutator of any element in a normal subgroup with an element in the whole group lies in the normal subgroup. This is one of the equivalent definitions of normal subgroup.
3	$\!c_{g}(a)=[g,a]a$ and $\!c_{g}(a+1)=[g,a+1](a+1)$		Step (1)	[SHOW MORE] Just follows from definitions: $[g,a]=gag^{-1}a^{-1}$ and $c_{g}(a)=gag^{-1}$. The same logic applies replacing $a$ by $a+1$.	The multiplicative commutator is not convenient because it is not additive/linear in either variable. So, we rewrite it in terms of $c_{g}$, which preserves the additive structure.
4	$c_{g}(a+1)=c_{g}(a)+1$		Step (1)	[SHOW MORE] Just follows from definitions: $c_{g}(a+1)=g(a+1)g^{-1}=gag^{-1}+g1g^{-1}=gag^{-1}+1$, which is $c_{g}(a)+1$.	More manipulation.
5	$\=[g,a]a+1$, or equivalently, $\!([g,a+1]-[g,a])a=1-[g,a+1]$		Steps (3) and (4)	[SHOW MORE] We plug in expressions for $c_{g}(a+1)$ and $c_{g}(a)$ from step (3) into the expression for step (4).	More manipulation.
6	The assumption $\![g,a+1]\neq [g,a]$ would lead to a contradiction, hence we must have $\![g,a+1]=[g,a]$.	$a\notin K$. Also, every nonzero element is invertible because $L$ is a skew field.	Steps (2), (5)	[SHOW MORE] If $[g,a+1]\neq [g,a]$, then the difference $[g,a+1]-[g,a]$ is invertible. Multiplying the second formulation of step (5) on the left by the inverse of $[g,a+1]-[g,a]$ and simplifying, we get $\!a=([g,a+1]-[g,a])^{-1}(1-[g,a+1])$. Both $[g,a+1]$ and $[g,a]$ are in $K^{*}$ by step (2). Thus, the right side is an expression involving terms in $K$ and hence must be in $K$. This contradicts the assumption that $a\notin K$.	More manipulation.
7	Plugging $[g,a+1]=[g,a]$ in the result of step (5) gives $[g,a]=1$.		Step (5)	[SHOW MORE] We get $[g,a](a+1)=[g,a]a+1$. Cancel out $[g,a]a$ additively form both sides, and we get $[g,a]=1$.

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 $h$.