I found myself reading Elements of Abstract Algebra by Allan Clark. This book has made me love math again. I like exercise $28\delta$.

$28\delta$. Let $S$ be a set with an operation which assigns to each ordered pair $(a, b)$ of elements of $S$ an element $a/b$ of $S$ in such a way that:

**P1.** there is an element $1\in S$, such that $a/b = 1$ if and
only if $a = b$;

**P2.** for any elements $a, b, c \in S$, $(a/c) / (b/c) = a/b$.

Show that $S$ is a group under the product defined by $ab = a/(1/b)$. (p. 21)

I'll start with three lemmas.

**Lemma 1.** $1/(a/b)$ is $b/a$.

$\begin{aligned}
1/(a/b) &= (b/b)/(a/b) &(\text{by P1.}) \\
&= b/a &(\text{by P2.})
\end{aligned}$

**Lemma 2.** Suppose $a/b=a/c$, then $b=c$.

$\begin{aligned}
a/b &= a/c \\
(a/b)/(1/c) &= (a/c)/(1/c) \\
(a/b)/(1/c) &= a/1 &(\text{by P2.}) \\
b &= c &(\text{by P2.})
\end{aligned}$

**Lemma 3.** $a/1$ is $a$.

$\begin{aligned}
1/(a/1) &= 1/a &(\text{by Lemma 1.}) \\
a/1 &= a &(\text{by Lemma 2.})
\end{aligned}$

Under the product $ab=a/(1/b)$, the **identity** is $1$ as $a1$ is
$a/(1/1)$ is $a/1$ by **P1**, is $a$ by **Lemma 3**. And $x$
**inverse** is $1/x$ as $x(1/x)$ is $x/(1/(1/x))$ is $x/(x/1)$ by
**Lemma 1**, is $1$ by **Lemma 3** and **P2**.

To show the product is **associative**, note $(ab)c=(a/(1/b))/(1/c)$
and $a(bc)=a/(1/(b/(1/c)))$. There are few steps to show these are the
same.

$\begin{aligned}
a(bc) &= a/(1/(b/(1/c))) \\
&= a/((1/c)/b) &(\text{by Lemma 1.}) \\
&= (a/(1/b))/(((1/c)/b)/(1/b)) &(\text{by P2.}) \\
&= (a/(1/b))/((1/c)/1) &(\text{by P2.}) \\
&= (a/(1/b))/(1/c) &(\text{by Lemma 3.}) \\
&= (ab)c
\end{aligned}$

So the product is associative and $S$ is a group under the product defined by $ab = a/(1/b)$.