Since the semigroup \(S\) is finite, for any \(x \in S\), the set of its powers \(\{ x^1, x^2, x^3, \dots \} \) must also be finite. By the pigeonhole principle, there must exist distinct positive integers \(m, n\) with \(m {\lt} n\) such that \(x^m = x^n\). This gives us \(x^m = x^m \cdot x^{n-m}\), showing that powers of \(x\) eventually enter a cycle of length \(\ell = n - m\) after an initial “tail” of length \(m\).

To construct an idempotent, we choose the exponent \(k = 2m\ell \). This exponent is chosen to satisfy two key properties: (1) it is beyond the pre-period \(m\), ensuring we are in the cyclic part, and (2) it is a multiple of the cycle length \(\ell \). Since \(k {\gt} m\) and adding multiples of \(\ell \) to an exponent beyond \(m\) does not change the result, we have:

where the last equality follows from the loop property applied with \(2m\) additional cycles of length \(\ell \). Thus \(x^k\) is idempotent.

For uniqueness, suppose \(x^a\) and \(x^b\) are both idempotent. Since idempotent elements satisfy \(e^n = e\) for all positive \(n\), we have:

Here we use that \(x^a\) raised to the power \(b\) equals \(x^{ab}\) (by associativity and the idempotent property), and similarly for \(x^b\) raised to the power \(a\). Since both expressions equal \(x^{ab}\), the two idempotent powers must be equal.

From \(a = xay\), we can repeatedly substitute \(a\) into itself to get \(a = x^k a y^k\) for any \(k \geq 1\). Since \(M\) is a finite monoid, there exists a non-zero power \(n_1\) such that \(x^{n_1}\) is an idempotent (by 1). Then we have \(a = x^{n_1} a y^{n_1}\). Multiplying by \(x^{n_1}\) on the left gives \(x^{n_1}a = x^{2n_1} a y^{n_1}\). Since \(x^{n_1}\) is idempotent, \(x^{2n_1} = x^{n_1}\), so \(x^{n_1}a = x^{n_1} a y^{n_1} = a\). A symmetric argument shows that \(ay^{n_2} = a\) for some idempotent power \(y^{n_2}\).