This question is inspired by this one:


[Can you do math without knowing how to count?](https://mathoverflow.net/questions/390427/can-you-do-math-without-knowing-how-to-count)


Let $M\_2$ be the set of words constructed by concatenation of the letters $a\_1$ and $a\_2$, with :


(\*) : for any $x$ word of $M\_2$ $xx = x$.


Is it true $card(M\_2)=card(\mathbb N) $?


If not, is it true $\exists n \in \mathbb N, card(M\_n)=card(\mathbb N) $?


The condition (\*) comes from the hypothesis that we assume that we do not know how to count.



Is the number of words finite, when you don't know how to count?
I'm upgrading my [comment](https://mathoverflow.net/questions/390668/is-the-number-of-words-finite-when-you-dont-know-how-to-count#comment995922_390668) to an answer, because I've found the source for the result asserted in the comment: For every $n$, the semigroup $M\_n$, presented by $n$ generators subject to the relations $xx=x$ for all words $x$, is finite. The reference is


Green, J. A.; Rees, D.
[On semi-groups in which x^r=x](https://doi.org/10.1017/s0305004100027341).
Proc. Cambridge Philos. Soc. 48 (1952), 35–40,


and its Mathematical Reviews number is [MR0046353](https://mathscinet.ams.org/mathscinet-getitem?mr=46353).


