Result due to Cohn, unique division ring whose unit group is a given group?

Solving the exercise

Uniqueness is easy, since addition is determined by the group operation on $G$ and the map $\phi$

$$x+y = x(1-(-x^{-1}y)) = x \phi(g x^{-1} y )$$

as long as $g x^{-1} y \neq 1$, but $g x^{-1} y = 1 \implies -x^{-1} y = 1 \implies y = -x$ so that $x+y= 0$.

Existence is either straightforward or false: verify this definition of $+$ defines an associative, abelian binary operation on $G \cup \{0\}$ with additive inverse $x\mapsto gx$, and that the operation distributes over the multiplication of $G$.

Lemma: $\phi$ is injective. [Property 2 shows $\phi$ is its own inverse]

Lemma: g has order dividing 2. [$\phi(g^{-1}) \stackrel{3}{=} g \phi(g) g^{-1} \stackrel{1}{=} \phi(g) \stackrel{2}{\implies} g^{-1} =g$]

Lemma: If x and y commute, then so do $\phi(x)$ and y. [$y\phi(x) = y\phi(x) y^{-1} y \stackrel{1}{=} \phi(yxy^{-1}) y = \phi(x) y$]

Lemma: g is central. $\phi(x) \stackrel{3}{=} g\phi(x^{-1}) x \stackrel{L}{=} gx\phi(x^{-1}) \stackrel{3}{=} gxg \phi(x) x^{-1} \stackrel{L}{=} gxgx^{-1} \phi(x) \implies gxgx^{-1}=1 \implies gx=xg$.

Abelian: Mostly easy, once you know you can move g around. $$x \phi(g x^{-1} y ) = x \phi( (y^{-1} x g)^{-1} ) = xg\phi( y^{-1} x g ) g x^{-1} y = x \phi( g y^{-1} x ) x^{-1} y = y (y^{-1}x \phi( gy^{-1}x) (y^{-1}x)^{-1} = y \phi( (y^{-1}x) (gy^{-1}x) (y^{-1}x)^{-1} ) = y\phi(gy^{-1}x) $$

Associative: This one is awful, and I think needs something like the fourth property to continue. $x+(y+z) = x+(y\phi(gy^{-1}z)) = x\phi(gx^{-1}y\phi(gy^{-1}z))$ $(x+y)+z = (x\phi(gx^{-1}y)) + z = x\phi(gx^{-1}y)\phi(g (x\phi(gx^{-1}y))^{-1} z )$ so we need to prove that: $$\phi(gx^{-1}y\phi(gy^{-1}z)) \stackrel{?}{=} \phi(gx^{-1}y)\phi(g (x\phi(gx^{-1}y))^{-1} z ) $$

Distributive: These are easy. The second one uses that g is central. $xy+xz = xy\phi(g(xy)^{-1} xz) = xy\phi(gy^{-1}x^{-1}xz) = x(y\phi(gy^{-1}z) = x(y+z)$ $yx+zx = yx\phi(g(yx)^{-1} zx) \stackrel{1}{=} y\phi( xgx^{-1}y^{-1}zxx^{-1}) x = y\phi(gy^{-1} z) x = (y+z)x$

Understanding the result

A division ring is uniquely determined by its group of units, the value of $-1$, and the map $\phi:x \mapsto 1-x$. Obviously there are restrictions on the group, the elements that could be $-1$, and the map $\phi$. The first three properties are all necessary.

Cohn studied identities that must be satisfied in skew fields (expanding work of Malcev) and actually wrote down some of them. Presumably some property similar to the fourth was sufficient.

The result is trivial or wrong

Suppose such a G and D exists. Then $\phi$ is defined on $D\setminus\{0,1\}$. The fourth property becomes $$1-x/y = (1-(1-x)(1-1/y))(1-1/y).$$

Suppose $x \notin \{0,1,\tfrac12\}$ and set $y=1-x$. Clearly $y \notin \{0,1,x\}$, so we can apply the fourth property to see that $x^2-x+1 = 0$, and in particular that D has at most 5 elements. Since the polynomial has no roots in a field of five elements, in fact the only division rings that work are the finite fields of size 2 and 3 [where the fourth property is vacuous] and the finite field of size 4 [where the fourth property is satisfied].

To make this exercise from Jacobson's Basic Algebra I correct, replace the last condition on the function $\varphi$ with the following one:

• $y\varphi(y^{-1}\varphi(x))=x\varphi(x^{-1}\varphi(y))$ for $x,y\in H, x\neq \varphi(y)$ (and so $y \neq \varphi(x)$).

This is saying, in effect, that $y-(1-x) = x-(1-y)$, which, when combined with the other assumptions, is enough to imply associativity of addition. I'm guessing that this--or something equivalent to it--is what Cohn stated and Jacobson intended.