# more on division in groups

In the parent entry, it is shown that a non-empty 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. 1.

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

2. 2.

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

3. 3.

(multiplication) $x\cdot y:=x/y^{-1}$ (we also write $xy$ for $x\cdot y$ for simplicity)

###### Theorem 1.

Let $G$ be a non-empty set with a binary operation $/$ on it such that

1. 1.

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

2. 2.

$(x/x)/((y/y)/y)=y$

hold for all $x,y,z\in G$. 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. 1.

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

2. 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

 $\displaystyle(xy)z$ $\displaystyle=$ $\displaystyle(xy)/z^{-1}=(x/y^{-1})/z^{-1}=(x/y^{-1})/((z^{-1}y^{-1})/y^{-1})$ $\displaystyle=$ $\displaystyle x/(z^{-1}y^{-1})=x/(yz)^{-1}=x(yz),$

completing the proof. ∎

There is also a companion theorem for abelian groups:

###### Theorem 2.

Let $G$ be a non-empty set with a binary operation $/$ on it such that

1. 1.

$x/(x/y)=y$

2. 2.

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

hold for all $x,y,z\in G$. 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. 1.

$x/e=x/(x/x)=x$

2. 2.

$e^{-1}=e/e=e$

3. 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 $xx^{-1}=x/(x^{-1})^{-1}=x/x=e$, 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 non-empty 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. 215-221, (1952).
Title more on division in groups MoreOnDivisionInGroups 2013-03-22 17:38:22 2013-03-22 17:38:22 CWoo (3771) CWoo (3771) 8 CWoo (3771) Result msc 08A99 msc 20A05 msc 20-00 AlternativeDefinitionOfGroup