0

Let $(M, \cdot, e)$ be a finite monoid. For $a \in M$ and some integer $n$, show there exists an integer $m >n$ such that $a^n = a^m$.

I've seen this conclusion used in many other questions, such as here and here, and I understand that the set $\lbrace e, a, a^2, ...\rbrace$ is finite, but I am still confused on how this explains the existence of the integer $m$.

Shaun
  • 47,747
RMG
  • 19
  • 2
  • Such application of the Pigeonhole principle of the set of powers in a finite monoid occurs here in many places, e.g. see the linked dupes (and their links). – Bill Dubuque Jul 08 '22 at 19:59
  • 2
    Those duplicates are for groups only, @BillDubuque. – Shaun Jul 08 '22 at 20:19
  • @Shaun Not true. The linked dupes use exactly the same pigeonhole argument as here. The argument is applied in many contexts (monoids, (semi)groups, rings, etc) but it is exactly the same argument in all cases - see abstract duplicates – Bill Dubuque Jul 08 '22 at 20:53
  • 1
    @BillDubuque: not directly related to the discussion here though, do you consider one should close all but one question under the tag partial-fractions as "duplicate" since they are essential one single abstract technique? –  Jul 08 '22 at 21:00
  • 1
    @user1046533 Dupe evaluation depends on the question and the potential dupe targets that currently exist. But this is not the proper place to discuss such meta matters., – Bill Dubuque Jul 08 '22 at 21:11

1 Answers1

-2

This follows from the Pigeonhole Principle: since there is infinitely many potential powers of $a$, each of which in the monoid (by closure), but only finitely many elements of the monoid, there must exist at least two powers of $a$ that are the same, simply because there is no other place for the infinitely many potential powers to go.

The pigeons are the potential powers of $a$; the pigeonholes are the elements of the finite monoid.

Shaun
  • 47,747
  • Is there any way to state this formally? for example if I have $x^n$, is there some $k$ such that $x^n = x^{n+k}$? – RMG Jul 08 '22 at 19:27
  • Yes, @RMG. Use the statement of the Pigeonhole Principle. – Shaun Jul 08 '22 at 19:28
  • Hang on, @RMG. There exist some finite monoids for which there exists a power $x^a$ of an element $x$ such that $x^a\neq x^{a+b}$ for all $b\in\Bbb N$; they're finite, cyclic monoids that are not groups. See here for details. – Shaun Jul 08 '22 at 20:00
  • @RMC They're also called monogenic semigroups with a one and denoted $M(m,r)^1$, were $m$ is the smallest natural number such that the generator to the $m+r$ is the same are that to the $m$, where $r$ is the smallest natural number. Note that $M(1,q)^1$ is simply the cyclic group of order $q$. – Shaun Jul 08 '22 at 20:12
  • 1
    @Shaun why this answer get $-3$ Downvote? – SoG Jul 25 '22 at 16:13
  • Supposedly, the question has a duplicate, @LostinSpace. – Shaun Jul 25 '22 at 16:15
  • 1
    @Shaun Ok. Sir. – SoG Jul 25 '22 at 16:17