Element has Idempotent Power in Finite Semigroup

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Theorem

Let $\left({S, \circ}\right)$ be a finite semigroup.

For every element in $\left({S, \circ}\right)$, there is a power of that element which is idempotent.


That is:

$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$


Proof

From Finite Semigroup Equal Elements for Different Powers, we have:

$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$.


Let $m > n$. Let $n = k, m = k + l$.

Then $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$.


Now we show that $x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$, i.e. that $x^{k l}$ is idempotent.

First:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x^k\) \(=\) \(\displaystyle \) \(\displaystyle x^{k + l}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^k \circ x^l\) \(=\) \(\displaystyle \) \(\displaystyle x^{k + l} \circ x^l\) \(\displaystyle \) \(\displaystyle \)          Both sides ${} \circ x^l$          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^{k + l}\) \(=\) \(\displaystyle \) \(\displaystyle x^{k + 2 l}\) \(\displaystyle \) \(\displaystyle \)          Index Laws for Semigroups‎          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^k\) \(=\) \(\displaystyle \) \(\displaystyle x^{k + 2 l}\) \(\displaystyle \) \(\displaystyle \)          as $x^k = x^{k + l}$          


From here we can easily prove by induction that $\forall n \in \N: x^k = x^{k + n l}$.


In particular, $x^k = x^{k + k l} = x^{k \left({l + 1}\right)}$.


There are two cases to consider:

$(1): \quad$ If $l = 1$, then $x^k = x^{k \left({l + 1}\right)} = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
$(2): \quad$ If $l > 1$, then:
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle x^k\) \(=\) \(\displaystyle \) \(\displaystyle x^{k + kl}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle x^{k \left({l + 1}\right)}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^k \circ x^{k \left({l - 1}\right)}\) \(=\) \(\displaystyle \) \(\displaystyle x^{k \left({l + 1}\right)} \circ x^{k \left({l - 1}\right)}\) \(\displaystyle \) \(\displaystyle \)          Both sides ${} \circ x^{k \left({l - 1}\right)}$          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^{k + k l - k}\) \(=\) \(\displaystyle \) \(\displaystyle x^{kl + k + k l - k}\) \(\displaystyle \) \(\displaystyle \)          Index Laws for Semigroups‎          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle x^{k l}\) \(=\) \(\displaystyle \) \(\displaystyle x^{k l + k l}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle x^{k l} \circ x^{k l}\) \(\displaystyle \) \(\displaystyle \)                    

... and again, $x^{k l}$ is idempotent.

$\blacksquare$


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