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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.

Martin Brandenburg
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7 Answers7

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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 $t > s$.

Mikko Korhonen
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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 $a=b^{k-1}$.

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

Finite Semigroup: Exists Idempotent Power

Theorem

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).

Community
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amWhy
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  • 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"). – amWhy Apr 06 '13 at 16:20
  • I do not agree with "stronger, and more general". Because it's a simple consequence of the question. –  Apr 07 '13 at 17:53
  • @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. – amWhy Apr 07 '13 at 18:06
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    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 $k$. –  Apr 07 '13 at 18:19
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    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. – amWhy Apr 07 '13 at 18:24
  • Actually, CutieKrait, I think we're splitting hairs. Your observation (in your post) was astute, and the answer you accepted provides the essentials of the proof (which happens to be the proof to which I linked you), with the added observation that we can obtain an idempotent element in any finite semigroup by equating it with the idempotent power of another element. – amWhy Apr 07 '13 at 18:35
  • thanks. my point was what you edited out! in fact I wanted to check if the theorem in your post is a corollary of the theorem in the book. –  Apr 07 '13 at 18:40
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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.)

Martin Sleziak
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Andrés E. Caicedo
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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

Martin Sleziak
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Mikasa
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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.

J.-E. Pin
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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 $$ (x^{jk})^2=x^{jk-i}\,x^{i+jk}=x^{jk-i}\,x^i=x^{jk}. $$

Pierre-Yves Gaillard
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