I am trying to understand the proof that the group of left quotients of a right reversible, cancellative semigroup is unique up to isomorphism, which can be found in Clifford's $`$The Algebraic Theory of Semigroups':
Clifford, A. H.; Preston, G. B., The algebraic theory of semigroups. Vol. I., Mathematical Surveys. 7. Providence, R.I.: American Mathematical Society (AMS). xv, 224 p. (1961). ZBL0111.03403.
However some steps are skipped in the proof and I'm struggling to connect each step. Here's what I have so far:
We want to find an isomorphism between two groups of left quotients of S, call them $(G,\cdot)$ and $(G', \circ)$. Elements in these groups are of the form $a^{-1}b$ and $a'^{-1}\circ b'$ respectively, where $a,b,a',b'\in S$. Hence, we want an isomorphism mapping $a^{-1}b\mapsto a^{-1}\circ b$. We essentially have the form of isomorphism that we need - we now need to verify it is indeed an isomorphism.
Now, we can express any element of the form $ab^{-1}$ in $G$ as $x^{-1}y$, for some $x,y\in S$. This implies that $xa=yb$. Clifford then states that for $ab^{-1}, cd^{-1}$ in $G$, we have that $ab^{-1}=cd^{-1}\iff (xa=yc \iff xb=yd)$, for $a,b,c,d\in S$.
Problem (1): I'm not sure why this is relevant to showing that our map is an isomorphism, and
Problem (2): I'm not sure how to prove this statement.
Can anyone give any hints as to how to address these problems?
