more on division in groups
In the parent entry, it is shown that a nonempty set $G$ equipped a binary operation^{} “$/$” called “division” satisfying three identities^{} has the structure^{} of a group. In this entry, we show that two identities are enough. Associated with every $x,y\in G$, we set

1.
$E(x):=x/x$,

2.
(inverse^{}) ${x}^{1}:=E(x)/x$, and

3.
(multiplication^{}) $x\cdot y:=x/{y}^{1}$ (we also write $xy$ for $x\cdot y$ for simplicity)
Theorem 1.
Let $G$ be a nonempty set with a binary operation $\mathrm{/}$ on it such that

1.
$(x/z)/(y/z)=x/y$

2.
$(x/x)/((y/y)/y)=y$
hold for all $x\mathrm{,}y\mathrm{,}z\mathrm{\in}G$. Then $G$ has the structure of a group
Proof.
From 1, we have $E(x/z)=(x/z)/(x/z)=x/x=E(x)$, so $E(E(x))=E(x/x)=E(x)$. From 2, we have $y=E(x)/{y}^{1}$, so $E(y)=E(E(x)/{y}^{1})=E(E(x))=E(x)$. This shows that $E:G\to G$ is a constant function, whose value we denote by $e$.
Note that $x=e/{x}^{1}$ by rewriting condition 2. This implies that $e\cdot x=e/{x}^{1}=x$. In addition, ${x}^{1}=e/x$ by rewriting the definition of the inverse. In particular, ${e}^{1}=e/e=e$. Furthermore, since $x/e=(e/{x}^{1})/({x}^{1}/{x}^{1})=e/{x}^{1}=x$, this implies that $x\cdot e=x/{e}^{1}=x/e=x$. So $e$ is the “identity” in $G$ with respect to $\cdot $.
Next, ${x}^{1}\cdot x={x}^{1}/{x}^{1}=e$. To see that $x\cdot {x}^{1}=e$, first observe that ${({x}^{1})}^{1}=e/{x}^{1}=x$, so $x\cdot {x}^{1}=x/{({x}^{1})}^{1}=x/e=x$. This shows that ${x}^{1}$ is the “inverse” of $x$ in $G$ with respect to $\cdot $.
Finally, we need to verify $(xy)z=x(yz)$. To see this, first note that

1.
$(xy)/y=(x/{y}^{1})/y=(x/{y}^{1})/(e/{y}^{1})=x/e=x$, and

2.
${(xy)}^{1}=e/(xy)=e/(x/{y}^{1})=({y}^{1}/{y}^{1})/(x/{y}^{1})={y}^{1}/x={y}^{1}{x}^{1}$.
From the two identities above, we deduce
$(xy)z$  $=$  $(xy)/{z}^{1}=(x/{y}^{1})/{z}^{1}=(x/{y}^{1})/(({z}^{1}{y}^{1})/{y}^{1})$  
$=$  $x/({z}^{1}{y}^{1})=x/{(yz)}^{1}=x(yz),$ 
completing the proof. ∎
There is also a companion theorem for abelian groups^{}:
Theorem 2.
Let $G$ be a nonempty set with a binary operation $\mathrm{/}$ on it such that

1.
$x/(x/y)=y$

2.
$(x/y)/z=(x/z)/y$
hold for all $x\mathrm{,}y\mathrm{,}z\mathrm{\in}G$. Then $G$ has the structure of an abelian group
Proof.
First, note that $E(x/y)=(x/y)/(x/y)=(x/(x/y))/y=y/y=E(y)$, so $E(x)=E((x/y)/x)=E((x/x)/y)=E(y)$, implying that $E$ is a constant function on $G$. Again, denote its value by $e$. Below are some simple consequences:

1.
$x/e=x/(x/x)=x$

2.
${e}^{1}=e/e=e$

3.
${({x}^{1})}^{1}={(e/x)}^{1}=e/(e/x)=x$
So, $xe=x/{e}^{1}=x/e=x$. Also, $ex=e/{x}^{1}=e/(e/x)=x$. This shows that $e$ is the “identity” of $G$ with respect to $\cdot $. In addition, ${x}^{1}x={x}^{1}/{x}^{1}=e$ and $x{x}^{1}=x/{({x}^{1})}^{1}=x/x=e$, showing that ${x}^{1}$ is the “inverse” of $x$ in $G$ with respect to $\cdot $.
Finally, we show that $\cdot $ is commutative^{} and associative. For commutativity, we have $xy=(ex)y=(e/{x}^{1})/{y}^{1}=(e/{y}^{1})/{x}^{1}=(ey)x=yx$, and associativity is shown by $x(yz)=(yz)x=(y/{z}^{1})/{x}^{1}=(y/{x}^{1})/{z}^{1}=(yx)z=(xy)z$. ∎
Remark. Remarkably, it can be shown (see reference) that a nonempty set $G$ with binary operation $/$ satisfying a single identity:
$$x/((((x/x)/y)/z)/(((x/x)/x)/z))=y$$ 
has the structure of a group, and satisfying
$$x/((y/z)/(y/x))=z$$ 
has the structure of an abelian group.
References
 1 G. Higman, B. H. Neumann Groups as groupoids^{} with one law. Publ. Math. Debrecen 2 pp. 215221, (1952).
Title  more on division in groups 

Canonical name  MoreOnDivisionInGroups 
Date of creation  20130322 17:38:22 
Last modified on  20130322 17:38:22 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  8 
Author  CWoo (3771) 
Entry type  Result 
Classification  msc 08A99 
Classification  msc 20A05 
Classification  msc 2000 
Related topic  AlternativeDefinitionOfGroup 