Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a simple question about *finite * monoids. Given a finite monoid M ( finite cardinality) given any element $$ a \in M $$ it it´s true that always exist an integer $n$ such that $ a^n = a $ ?

Someone has a site or a book that provides a lot of examples of finite monoids that are not groups?

share|improve this question

3 Answers 3

$n=1$ always works, but there may not be a larger $n$ that does.

Consider for example, multiplication modulo $4$, which is a finite monoid with elements $\{0,1,2,3\}$ and identity $1$. Then for $a=2$ we have $a^n=0\neq a$ for all $n\ge 2$.

share|improve this answer

I assume you want $n>1$.

In that case, the answer is no. For any positive integer $m$, the set $\{1, x, x^2, \ldots, x^m \}$ forms a monoid under the product $x^i * x^j = x^{i+j}$ if $i+j < m$, and $x^m$ otherwise, and (if $m>1$) then $x^n \ne x$ for all $n>1$.

share|improve this answer

The only thing I can tell is this : given $a \in M$, consider the sequence $$ \{a, a^2, a^4, a^8, a^{16}, \dots\} $$ Since the monoid is finite, at some point the sequence will repeat itself, hence $$ \exists k,j > 0 \text{ s.t. } a^{2^k} = a^{2^{k+j}}=a^{2^k 2^j} = (a^{2^k})^{2^j}. $$ This means that $b = a^{2^k}$ is such that $b^{2^j} = b$. This doesn't mean that it works for any $a$, but at least for some of them. (Note that some of them may be only the identity if, for instance, the monoid has the trivial multiplication on it which sends everything to the same element.)

Hope that helps,

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.