Exercise 24, § 4

(a) Given any set E, consider in the set $E \times E$ the law of composition, not everywhere defined, written multiplicatively, under which the composition of $(x, y)$ and $(y', z)$ is only defined if $y' = y$ and in this case has the value $(x, z)$. Show that $E \times E$, with this law of composition, is a groupoid (Exercise 23).

(b) Let $(x, y) \equiv (x', y')$ (R) be an equivalence relation compatible with the above composition (that is such that $(x, y) \equiv (x', y')$ and $(y, z) \equiv (y', z')$ imply $(x, z) \equiv (x', z')$); suppose further that the relation R satisfies the following condition: for all $x, y, z$ there exists one and only one $t \in E$ such that $(x, y) \equiv (z, t)$ and there exists a $u \in E$ such that $(x, y) \equiv (u, z)$.

Show that under these conditions the quotient structure by R of the groupoid structure on $E \times E$ is a group structure (prove first that the quotient law is everywhere defined, since, if $\dot{x}, \dot{y}, \dot{z}$ are three classes such that $\dot{x}\dot{y} = \dot{x}\dot{z}$, then $\dot{y} = \dot{z}$; establish finally that, for all $x \in E, y \in E, (x, x) \equiv (y, y)$).

(c) Let G be the group obtained by giving the quotient of $E \times E$ by R the above structure. Let $a$ be any element of E; if, for all $x \in E, f_a(x)$ denotes the class mod. R of $(a, x)$, show that $f_a$ is a bijective mapping of E onto G and that the relation $(x, y) \equiv (x', y')$ is equivalent to the relation

For G to be commutative, it is necessary and sufficient that the relation $(x, y) \equiv (x', y')$ imply $(x, x') \equiv (y, y')$.