I am taking a class on Semigroup Theory and I am really struggling to get my head around the material. I have to prove the following result, which seems to be a pretty standard result in Semigroup Theory:
Let $R^{\#}$ denote the smallest congruence containing the relation $R$ on $S$, that is, the congruence generated by $R$. Then $(x,y)\in R^{\#}$ if and only if either $x=y$ or there is a finite sequence $\{(x_i,y_i):i=1,\ldots,n\}\subseteq R$ and $\{c_i:i=1\ldots,n\}\subseteq S$ with $x=c_1x_1$, $c_ix_i=c_{i+1}y_{i+1}$ for $i=1,\ldots,n-1$, and $c_ny_n=y$.
Could anyone give me a hint with how to start this proof? Any help would be greatly appreciated. Thanks in advance!
Let $S$ be a nontivial monoid and let $R$ be the equality relation. Then $R^{\#} = R$. On the other hand, let $x$ and $y$ be two elements of $S$. Taking $c_1 = 1$, $x_1 = y_1 = x$, $c_2 = y$, $x_2 = y_2 = 1$, one has $x = c_1x_1$, $y = c_2y_2$ and $(x_1, y_1), (x_2, y_2) \in R$. Thus your statement is incorrect.
Another reason why your assertion looks suspicious is that you only use left multiplication on elements of $R$ and never use right multiplication.