Let $(G,\cdot)$ be a non-empty finite semigroup. Is there any $a\in G$ such that: $$a^2=a$$

It seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof?

Theorem 2.2.1. [R. Ellis] Let $S$ be a compact right topological semigroup. Then there exists an idempotent in it.

This theorem is also known as Ellis–Numakura lemma.

  • $\begingroup$ "Pages 77 to 99 are not shown in this preview". Do you have another link? $\endgroup$
    – Julien
    Apr 6, 2013 at 15:42
  • $\begingroup$ See also this post on MO and the comments there. $\endgroup$ Oct 19, 2015 at 10:46

7 Answers 7


Note first that it suffices to prove that $a^k = a$ for some $k \geq 2$. If $k = 2$ we are done. Otherwise $k > 2$ and multiplying both sides by $a^{k-2}$ gives $(a^{k-1})^2 = a^{k-1}$.

Fix $x \in G$ and consider the sequence

$$x, x^2, x^4, x^8, x^{16}, \ldots$$

Since $G$ is finite, there is repetition in this sequence. That is, $x^{2^t} = x^{2^s}$ for some integers $t > s \geq 1$. Thus $x^{2^t} = (x^{2^s})^{2^{t-s}} = x^{2^s}$, so choosing $a = x^{2^{s}}$ and $k = 2^{t-s}$ gives $a^k = a$. Note that $k \geq 2$ since $t > s$.

  • 3
    $\begingroup$ Excellent answer +1 $\endgroup$
    – Learnmore
    Jan 14, 2017 at 6:17

Pick an arbitrary element and start iterating $x\mapsto x^2$. Since the semigroup is finite you will eventually hit a cycle. This gives you a $b$ such that $b^k=b$ for some $k\ge 2$. Now set $a=b^{k-1}$.


You can find a proof $^1$ of the following theorem, from which your assertion follows, at Proof Wiki

Finite Semigroup: Exists Idempotent Power


Let $(S,\circ)$ be a finite semigroup.

For every element in $(S,\circ),$ there is a power of that element which is idempotent. That is: $$\forall x \in S:\exists i \in \mathbb N:x^i=x^i\circ x^i$$

Essentially, then, for your purposes: you can simply set $a = x^i$,
and you have the existence of an idempotent element $a \in S$ such that $\;a^2 = a$.

$1.$ Source of proof: Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978).

  • $\begingroup$ N.B. Proof Wiki is a great resource when looking for a proof, or an alternative proof, to mathematical proofs. (Link is to "main page"). $\endgroup$
    – amWhy
    Apr 6, 2013 at 16:20
  • $\begingroup$ I do not agree with "stronger, and more general". Because it's a simple consequence of the question. $\endgroup$
    – user59671
    Apr 7, 2013 at 17:53
  • $\begingroup$ @CutieKrait: Actually, no, this theorem talks about every element of a finite semigroup, not just the existence of some element such that... Your question and its answer follow from the theorem above. Besides, the point of my post was not to argue the merits of the theorem or your question, the point of my post was to simply offer help. $\endgroup$
    – amWhy
    Apr 7, 2013 at 18:06
  • 2
    $\begingroup$ assume we know every finite element in a semigroup $S$ has an idempotent. Then for each $a$ in $S$ , $\{a^n \mid n\in \Bbb N \}$ is a semigroup. So it has an idempotent, say $a^k$. $a^ka^k=a^k$ for some $k$. $\endgroup$
    – user59671
    Apr 7, 2013 at 18:19
  • 1
    $\begingroup$ You asked about whether there exists an element in each semigroup...such that it is idempotent. We don't know that for each x $x \circ x = x$, we only know (from the theorem) that for each $x$, exists a power $i$ of $x$ such that $x^i\circ x^i = x^i$...that's what the theorem above states. From that we know that exists $y = x^i$ such that $y\circ y = y$, we don't know that for all $a$ in the semigroup, $a\circ a = a$. The assumption that "every finite element in a semigroup has [a power that is] an idempotent" is NOT to say ever finite element in a semigroup is idempotent. $\endgroup$
    – amWhy
    Apr 7, 2013 at 18:24

When I first saw the question, I remembered there was a proof on MO using Ramsey theory, but couldn't remember how the argument went, so I came up with the following, that I first posted as a comment:

A cute proof using Schur's theorem: Fix $a$ in your semigroup $S$, and color $n$ and $m$ with the same color whenever $a^n=a^m$. By Schur's theorem (and the fact that the semigroup is finite) there are $n\le m$ such that $n$, $m$, and $n+m$ have the same color. That is, $a^n=a^m=a^{n+m}=(a^n)^2$.

(Today I finally found the thread on MO with the Ramsey theory proof, using Ramsey's theorem directly rather than Schur's theorem.)


This is a very good point proved by E.H.Moore that says:

Some power of every element of a finite semigroup is idempotent.

Trans. Amer. Math. Soc. 3 1902


Let $x$ be an element of $S$. Since $S$ is finite, there exist integers $i, p >0$ such that $x^i = x^{i+p}$. It follows that, for all $k \geqslant i$, $x^k = x^{k+p}$. In particular, if $k$ is a multiple of $p$, say $k = qp$, one has $(x^k)^2 = x^{2k} = x^{k+qp} = x^k$ and thus $x^k$ is idempotent.


This is essentially the J.-E. Pin's argument. Let $x$ be in your semigroup. There are positive integers $i,j$ such that $x^{i+j}=x^i$. We have $x^{i+jk}=x^i$ for all positive integer $k$ (induction on $k$). Choosing $k$ so that $jk>i$ we get $$ (x^{jk})^2=x^{jk-i}\,x^{i+jk}=x^{jk-i}\,x^i=x^{jk}. $$


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