I found myself reading Elements of Abstract Algebra by Allan Clark. This book has made me love math again. I like exercise 28δ28\delta.

28δ28\delta. Let SS be a set with an operation which assigns to each ordered pair (a,b)(a, b) of elements of SS an element a/ba/b of SS in such a way that:

P1. there is an element 1S1\in S, such that a/b=1a/b = 1 if and only if a=ba = b;

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

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

I'll start with three lemmas.

Lemma 1. 1/(a/b)1/(a/b) is b/ab/a.

1/(a/b)=(b/b)/(a/b)(by P1.)=b/a(by P2.)\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/ca/b=a/c, then b=cb=c.

a/b=a/c(a/b)/(1/c)=(a/c)/(1/c)(a/b)/(1/c)=a/1(by P2.)b=c(by P2.)\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/1a/1 is aa.

1/(a/1)=1/a(by Lemma 1.)a/1=a(by Lemma 2.)\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)ab=a/(1/b), the identity is 11 as a1a1 is a/(1/1)a/(1/1) is a/1a/1 by P1, is aa by Lemma 3. And xx inverse is 1/x1/x as x(1/x)x(1/x) is x/(1/(1/x))x/(1/(1/x)) is x/(x/1)x/(x/1) by Lemma 1, is 11 by Lemma 3 and P2.

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

a(bc)=a/(1/(b/(1/c)))=a/((1/c)/b)(by Lemma 1.)=(a/(1/b))/(((1/c)/b)/(1/b))(by P2.)=(a/(1/b))/((1/c)/1)(by P2.)=(a/(1/b))/(1/c)(by Lemma 3.)=(ab)c\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 SS is a group under the product defined by ab=a/(1/b)ab = a/(1/b).