Let $M\curvearrowright X$ be a monoid action. Say it's strongly transitive if it has only one strong orbits, i.e if $\forall x,y\exists m\in M:mx=y$.
Page 5 of this paper says that if $M$ is finite, then itself as an $M$-set is strongly transitive iff it's a group.
What goes wrong if $M$ isn't finite?
Added. More interestingly, let $f:A\to B$ be a monoid homomorphism. Let $\operatorname{Ker}f\curvearrowright A$ be the canonical multiplicative action of the kernel. Are the following equivalent?
- The connected components of $\operatorname{Ker}f\curvearrowright A$ are strongly connected;
- $A$ is a group.
Update. I realized it is completely unreasonable to expect the property of being a group to be captured by a single homomorphism out of it. I would like to update my question to the following:
For a monoid homomorphism $f:A\to B$ let $\operatorname{Ker}f\curvearrowright A$ be the canonical multiplicative action of the kernel. Are the following equivalent?
- The connected components of $\operatorname{Ker}f\curvearrowright A$ are strongly connected for all $f$;
- $A$ is a group.
It is ok for infinite monoids too. If $a\in M$, then $ba=1$ for some $b\in M$ and $cb=1$ for some $c\in M$. So $c=cba=a$ and hence $b$ and $a$ are inverses. My paper you quoted is about finite monoids but that doesn't mean that the assumption is always needed.
The answer to your second question is no. Take $B$ to be the monoid $0,1$ with multiplication and take $A$ to be the result of adjoining an identity $1$ to any non-empty semigroup. Take your homomorphism to take 1 to 1 and all other elements to 0.