Monoid is not Empty
Jump to navigation
Jump to search
Theorem
Proof
Let $\left({S, \circ}\right)$ be a the monoid.
By definition:
- Identity: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$
So a monoid must at least have an identity.
Therefore $e_S \in S$ and so $S$ is not the empty set.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids