I'm currently working on a problem where a nonempty set $S$ has $\cdot$ an associative and cancellable operation. It is a bit ambiguous but when it says operation I feel the problem is talking about a binary operation, then this set would be a cancellable semigroup.
$S$ also has the property that for every element the set of its powers is finite. Then, every element in $S$ has finite order, since the order is definided as the cardinality of such set.
The question is,
Should $S$ be group?
When reading up on this question, I found out that if $S$ is itself finite, then $(S,\cdot)$ is a group. I tried setting up a similar function $s\mapsto as$, but I couldn't find a right inverse given that the elements had finite order.
The thing is that, if every element has finite order, must there be an identity element? I feel like there would be a lot of identities, $s^0$. But I haven't found a way to prove uniqueness or that it even exists.
I also thought of a structure where all elements have finite order but is infinite and came about with $\Bbb Q/\Bbb Z$. Now, this set is that it is an infinite group where all elements have a finite order. The problem is, it's already a group.
I can't seem to find a counterexample or give a proof to the problem. Any hints or solutions are kindly appreciated.