Example

Let V be an A-module and S a monoid which operates on V on the left; this means (I, § 5, no. 1) that there is given a mapping $(s, x) \mapsto s.x$ of S into V such that $s.(x + y) = s.x + s.y, s.(\alpha x) = \alpha(s.x)$ and $s.(t.x) = (st).x$ for $s, t$ in S, $x, y$ in V and $\alpha \in A$ and, denoting by e the identity element of S, e.x = x for x ∈ V. Writing f(s)(x) = s.x, f is a homomorphism of S into the algebra End_A(V) (with only its multiplicative law), mapping the identity element e to the unit element l_V. Applying Proposition 6, an A-algebra homomorphism $\bar{f}: A^{(S)} \to \mathrm{End}_A(V)$ is obtained, which gives the underlying group of V a left module structure over $A^{(S)}$.

This allows us to reduce the study of commutative groups with operators to that of modules. For let M be a commutative group with operators written additively, all of whose external laws are written multiplicatively. Let $\Omega$ be the sum set (Set Theory, II, § 4, no. 8) of the domains of operators of the various external laws on M, each of these domains being canonically identified with a subset of $\Omega$. Let Mo($\Omega$) be the free monoid (I § 7, no. 2) constructed on $\Omega$; a law of action

is defined on M with Mo($\Omega$) as domain of operators, by induction on the length of the word s in Mo($\Omega$); if s is of length 0, it is the empty word e and we write $e.x = x$ for all $x \in M$. If x is of length $n \geq 1$, it can be written uniquely as $s = tu$, where u is of length $n - 1$ and t of length 1, so that $t \in \Omega$; we then write $s.x = t.(u.x)$. For any two words s, $s'$ in Mo($\Omega$), the relation $s.(s'.x) = (ss').x$ is verified by induction on the length of s.

Then applying the method described above, a left $\mathbf{Z}^{(Mo(\Omega))}$-module structure is obtained on M and it is verified without difficulty that the usual notions in the theory of groups with operators (stable subgroups, homomorphisms) are the same for commutative groups with operators and the modules thus associated with them.