Cartan-Brauer-Hua theorem

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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
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Let K < L be Division ring (?)s (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.

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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.