Let be a semigroup. Prove that the following are equivalent.
- For every , there exists a unique such that is idempotent.
- For every , there exists a unique element such that .
- is regular and contains exactly one idempotent.
- is a group.
We will follow the strategy .
: Suppose that for every , there exists a unique such that is idempotent. Let . Now there exists a unique such that . Note that , so that (by the uniqueness of ) we have . Now and , so that . Suppose now that for some . Then , and again by uniqueness, . So for all there exists a unique such that .
: Suppose that for all there exists a unique with . Now is regular by definition. Now , so that by uniqueness, . So is an inverse for . Now let be idempotent (these exist in since for example is idempotent) and let with inverse . Now for some , so that . Now , so that . So . Now . Since every element of is the inverse of some other element, in fact is a right identity. A similar argument shows that is a left identity. So every idempotent is an identity. Since a semigroup can have at most one identity, has a unique idempotent.
: Suppose is regular and has a unique idempotent . Now for any , there exists an element such that . Now and , so that and are idempotent. Thus . In particular, we have for all . So is an identity element, and every has an inverse with respect to . So is a group.
: Suppose is a group with identity . Denote the inverse of by . Certainly is idempotent. Now if is idempotent, we have , so that , so . So if is idempotent, then and thus . That is, for every there is a unique such that is idempotent.