Prove that in a finite monoid there exists $m > n$ such that $a^m = a^n$

236 Views Asked by At

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 $m$.

1

There are 1 best solutions below

7
On BEST ANSWER

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.