25
$\begingroup$

Let $(G,\cdot)$ be a 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.

$\endgroup$
  • $\begingroup$ "Pages 77 to 99 are not shown in this preview". Do you have another link? $\endgroup$ – Julien Apr 6 '13 at 15:42
  • $\begingroup$ See also this post on MO and the comments there. $\endgroup$ – Martin Sleziak Oct 19 '15 at 10:46
37
$\begingroup$

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 $t > s$.

$\endgroup$
  • 1
    $\begingroup$ Excellent answer +1 $\endgroup$ – Learnmore Jan 14 '17 at 6:17
21
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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.)

$\endgroup$
6
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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"). $\endgroup$ – Namaste Apr 6 '13 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 '13 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. $\endgroup$ – Namaste Apr 7 '13 at 18:06
  • 1
    $\begingroup$ 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 '13 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. $\endgroup$ – Namaste Apr 7 '13 at 18:24
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy