Let $S$ be a regular semigroup, $\phi:S \rightarrow T$ an epimorphism... Prove that $\phi(a)=c$

Question

Let $S$ be a regular semigroup, $\phi: S \rightarrow T$ onto morphism of semigroups, $c,d \in T$ mutually inverses, ie, $c=cdc$ and $d=dcd$.

Suppose that $c=\phi(x)$ and $d=\phi(y)$, where $x,y \in S$.

Let $v$ be an inverse of $xyxy$, ie, $v=vxyxyv$ and $xyxy=xyxyvxyxy$.

Let $a=xyvxyx$ and $b=yvxy$.

Prove that $\phi(a)=c$.

In all my attempts the element $v$ does not disappear, which is a big problem.

I also proved that $\phi(v)=\phi(v)cd\phi(v)$, so $\phi(a)=[cd\phi(v)]^{n}c$ for all $n \in \mathbb{N}$.

Can someone give me a hint or show me the trick to solve the problem?

Answer 1

Let $b = \phi(v)$. Then $\phi(v) = \phi(vxyxyv)$ and $\phi(xyxy) = \phi(xyxyvxyxy)$, that is, $b = b(cdcd)b$ and $cdcd = (cdcd)b(cdcd)$. Now, observing that $cd = cdcd$, we get $$ \phi(a) = \phi(xyvxyx) = cdbcdc = (cdcd)b(cdcd)c = (cdcd)c = cdc = c $$