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?
On
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 $n>1$.
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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,
$n=1$ 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 $n\ge 2$.