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Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

Moreover, if a 0-simple semigroup $S$ with 1 has no other idempotents, then it is a group with 0.

Proof: Let $G$ be its subgroup of invertible elements. For every $a\in S\setminus 0$ there are such $x,y\in S$ that $xay=1$. Then $(ayx)^2=ayx$ whence $ayx=1$. Since $xay=ayx=1$, hence $x\in G$. But then $a=x^{-1}y^{-1}\in G$, i.e. $S=G\cup 0$.

Any regular $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

Moreover, if a 0-simple semigroup $S$ with 1 has no other idempotents, then it is a group with 0.

Proof: Let $G$ be its subgroup of invertible elements. For every $a\in S\setminus 0$ there are such $x,y\in S$ that $xay=1$. Then $(ayx)^2=ayx$ whence $ayx=1$. Since $xay=ayx=1$, hence $x\in G$. But then $a=x^{-1}y^{-1}\in G$, i.e. $S=G\cup 0$.

Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

Moreover, if a 0-simple semigroup $S$ with 1 has no other idempotents, then it is a group with 0.

Proof: Let $G$ be its subgroup of invertible elements. For every $a\in S\setminus 0$ there are such $x,y\in S$ that $xay=1$. Then $(ayx)^2=ayx$ whence $ayx=1$. Since $xay=ayx=1$, hence $x\in G$. But then $a=x^{-1}y^{-1}\in G$, i.e. $S=G\cup 0$.

Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.

If a 0-simple semigroup is not completely 0-simple, nevertheless it can have an unique idempotent $\ne 0$, e.g. a bicyclic monoid.

Any $J$-class with a joined zero is a 0-simple semigroup. So your question is reduced to the following: who many idempotents has a 0-simple semigroup? In particular, if $S$ is finite, a $J$-class (with 0) is completely 0-simple semigroup, so it has just one idempotent $\ne 0$ iff it is a group.