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This is a result of Cohn I would like to understand.

Let $G$ be a group, $g\in G$, and $\varphi$ a map of the subset $H=\{x\in G:x\neq 1\}$ into itself which has the properties

- $\varphi(yxy^{-1})=y(\varphi x)y^{-1}$ for $x\in H$, $y\in G$.
- $\varphi^2(x)=x$
- $\varphi(x^{-1})=g(\varphi x)x^{-1}$
- $\varphi(xy^{-1})=(\varphi(\varphi(x)\varphi(y^{-1})))\varphi(y^{-1})$ for $x,y\in H, x\neq y$.

Then there exists a unique division ring $D$ such that $D^*=G$ for the set of nonzero elements $D^*$ of $D$, and in $G$, $\varphi x=1-x$, $x\in H$, $g=-1$.

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I did notice that the first three properties agree with $\varphi x=1-x$ for $x\in H$. However, for the fourth property to be satisfied, would require

$$

\begin{align*}

1-xy^{-1} &= (1-(1-x)(1-y^{-1}))(1-y^{-1})\\

&= x+y^{-1}-2xy^{-1}-y^{-2}+xy^{-2}

\end{align*}

$$

which seems like an odd relation to have.

I thought about this, but really had no idea where to begin. Is there a reference for this result, or even a sketch I could try to work through? Thank you.

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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.

$xy+xz = xy\phi(g(xy)^{-1} xz) = xy\phi(gy^{-1}x^{-1}xz) = x(y\phi(gy^{-1}z) = x(y+z)$

$yx+zx = yx\phi(g(yx)^{-1} zx) \stackrel{1}{=} y\phi( xgx^{-1}y^{-1}zxx^{-1}) x = y\phi(gy^{-1} z) x = (y+z)x$

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.

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].

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