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

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 non-empty 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 non-empty 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 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.