Equivalent characterizations of groups

Let S be a semigroup. Prove that the following are equivalent.

  1. For every a \in S, there exists a unique x \in S such that ax is idempotent.
  2. For every a \in S, there exists a unique element x \in S such that a = axa.
  3. S is regular and contains exactly one idempotent.
  4. S is a group.

We will follow the strategy (1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4).

(1) \Rightarrow (2): Suppose that for every a \in S, there exists a unique x \in S such that ax is idempotent. Let a \in S. Now there exists a unique x \in S such that axax = ax. Note that a(xax)a(xax) = a(xax), so that (by the uniqueness of x) we have x = xax. Now xaxa = xa and x(axa)x(axa) = x(axa), so that a = xax. Suppose now that a = aya for some y \in S. Then ayay = ay, and again by uniqueness, y = x. So for all a \in S there exists a unique x \in S such that a = axa.

(2) \Rightarrow (3): Suppose that for all a \in S there exists a unique x \in S with a = axa. Now S is regular by definition. Now a(xax)a = axa = a, so that by uniqueness, x = xax. So x is an inverse for a. Now let e \in S be idempotent (these exist in S since for example ax is idempotent) and let a \in S with inverse x. Now ea = eayea for some y \in S, so that y = yeay. Now ye = yeaye, so that a = ayea. So ye = x. Now xe = ye^2 = ye = x. Since every element of S is the inverse of some other element, in fact e is a right identity. A similar argument shows that e is a left identity. So every idempotent is an identity. Since a semigroup can have at most one identity, S has a unique idempotent.

(3) \Rightarrow (4): Suppose S is regular and has a unique idempotent e. Now for any a \in S, there exists an element x \in S such that a = axa. Now axax = ax and xaxa = xa, so that xa and ax are idempotent. Thus ax = xa = e. In particular, we have ea = ae = a for all a \in S. So e \in S is an identity element, and every a \in S has an inverse with respect to e. So S is a group.

(4) \Rightarrow (1): Suppose S is a group with identity e. Denote the inverse of a by a^{-1}. Certainly aa^{-1} = e is idempotent. Now if x \in S is idempotent, we have xx = x, so that xx^{-1} = xx^{-1}, so x = e. So if ab = e is idempotent, then a^{-1}ab = a^{-1} and thus b = a^{-1}. That is, for every a \in S there is a unique x \in S such that ax is idempotent.

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