Exercise 28d.

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:
(1) there is an element $1 \in S$, such that $a/b=1$ if and only if $a = b$; (2) 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 $ab = a/(1/b)$.

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Proof

As usual, we need to show that the four group axioms hold for $S$.

1. Closure: Let $x, y \in S$. We need to show that $ab \in S$, that is, that $a/(1/b) \in S$. It will suffice to show that $1/b \in S$. We know that $1 \in S$, (1) tells us that. We also know, that for each ordered pair $a, b$ of elements in $S$, $a/b \in S$ as well. Now, for $a = 1$ and $b = b$^[1], we get that $1/b \in S$. Which is, as we said, enough for $ab$ to be in $S$.
2. Associativity: Let $a, b, c \in S$. We need to show that $(ab)c = a(bc)$. We have $$(i) \quad (ab)c = (a/(1/b))c=(a/(1/b))/(1/c)$$ and $$(ii) \quad a(bc) = a(b/(1/c)) = a/(1/(b/(1/c)))$$

Let's consider the "denominator" in $(ii)$ first. Note that we can rewrite the $1$ in its "numerator" as $(1/c)/(1/c)$, given the first assumption of the exercise. Using this we can then rewrite $1/(b/(1/c))$ as $((1/c)/(1/c))/(b/(1/c))$. Now we can use the second assumption of the exercise, and cancel out the "denominators" of the big "fraction" — we get $(1/c)/b$. Plugging this result back in $(ii)$ we get $a/((1/c)/b)$.

Let's now consider the "numerator" in $(i)$. unfinished

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Additional thoughts