Remark

(1) Theorem 1 can also be expressed by saying that $(E_S, \varepsilon)$ is the solution of the universal mapping problem for E, relative to monoids, monoid homomorphisms and homomorphisms of E into monoids which map the elements of S to invertible elements (Set Theory, IV, , no. 1). It follows (loc. cit.) that every other solution of this problem is isomorphic in a unique way to $(E_S, \varepsilon)$

(2) For the existence of a solution to the above universal mapping problem it is unnecessary to assume that the monoid E is commutative, as follows from Set Theory, IV, § 3, no. 2 (cf. Exercise 17).

We mention two important special cases of monoids of fractions.

(a) Let $\overline{E} = E_E$. As the monoid $\overline{E}$ is generated by the set $\varepsilon(E) \cup \varepsilon(E)^*$, which is composed of invertible elements, every element of $\overline{E}$ is invertible (no. 3, Corollary 2 to Proposition 4). In other words, $\overline{E}$ is a commutative group. Moreover, by Theorem 1 every homomorphism $f$ of E into a group G can be uniquely factorized in the form $f = \bar{f} \circ \varepsilon$, where $\bar{f}: \overline{E} \to G$ is a homomorphism. $\overline{E}$ is called the group of fractions of E (or group of differences of E in the case of additive notation).

(b) Let $\Phi = E_{\Sigma}$, where $\Sigma$ consists of the cancellable elements of E. By Proposition 6, (ii), the canonical homomorphism of E into $\Phi$ is injective; it will be profitable to identify E with its image in $\Phi$. Hence E is a submonoid of $\Phi$, every cancellable element of E has an inverse in $\Phi$ and every element of $\Phi$ is of the form $a/p = a.p^*$ with $a \in E$ and $p \in \Sigma$; then $a/p = a'/p'$ if and only if $ap' = pa'$. It is easily seen that the invertible elements of $\Phi$ are the fractions $a/p$, where $a$ and $p$ are cancellable, and $p/a$ is the inverse of $a/p$.

Now let S be a set of cancellable elements of E and $S'$ be the submonoid of E generated by S. If $a/p$ and $a'/p'$ are two elements of $E_S$, then $a/p = a'/p'$ if and only if $ap' = pa'$ (for $sap' = spa'$ implies $ap' = pa'$ for all $s \in S'$). $E_S$ may therefore be identified with the submonoid of $\Phi$ generated by $E \cup S^*$.

If every element of E is cancellable, then $\Phi = \overline{E}$ and E is a submonoid of the commutative group $\Phi$. Conversely, if E is isomorphic to a submonoid of a group, every element of E is cancellable.