Let $S$ be an inverse semigroup and consider the relation $\sigma$ on $S$ given by $$a \sigma b \iff ab^{-1} \in E(S)$$ Consider the congruence generated by $\sigma$, say $\sigma^*$.
Prove that $\sigma^*$ is the least group congruence on $S$.
That $\sigma^*$ is a congruence it is.
I don't know how to prove that $S/\sigma^*$ is a group and that if $\rho$ is a group congruence, $\sigma^* \subseteq \rho$.
Can someone give me a hint? Thanks!
Towards showing that $S/\sigma^*$ is a group: can you show that (the equivalence class corresponding to) $aa^{-1}$ is an identity in $S/\sigma^*$? More precisely: that $baa^{-1}\sigma b$ for any $b$?
Towards showing that $\sigma^*$ is the least group congruence: suppose $S/\rho$ is a group. Then show that if $a\sigma b$, the equivalence class corresponding to $ab^{-1}$ is an idempotent in $S/\rho$. But what are the idempotents in a group?