Percisely

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 $ab = a/(1/b)$ .