# 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.

## 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] 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]
3 $\! c_g(a) = [g,a]a$ and $\! c_g(a + 1) = [g,a+1](a + 1)$ Step (1) [SHOW MORE] 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] More manipulation.
5 $\! [g,a+1](a+1) = [g,a]a + 1$, or equivalently, $\! ([g,a+1] - [g,a])a = 1 - [g,a+1]$ Steps (3) and (4) [SHOW MORE] More manipulation.
6 The assumption $\! [g,a+1] \ne [g,a]$ 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] More manipulation.
7 Plugging $[g,a+1] = [g,a]$ in the result of step (5) gives $[g,a] = 1$. Step (5) [SHOW MORE]

### 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 $h$.