Question on proving $γ∨ρ=γ∘ρ∘γ$

Question

in theorem 5 of Group congruences on eventually regular semigroups by S. Hanumantha Rao. he saied it suffices to prove $ρ∘γ∘ρ⊆γ∘ρ∘γ$. Why?

Answer 1

I don't know what is your general definition for the join of two congruences, but in any algebra the join of two congruences $\gamma$ and $\rho$ (or indeed, in any set the join of those equivalence relations) is given by $$\gamma \vee \rho = \bigcup_{n\geq 1}(\gamma \circ \rho)^n.$$ Now, if $\rho \circ \gamma \circ \rho \subseteq \gamma \circ \rho \circ \gamma$, then \begin{align} (\gamma \circ \rho \circ \gamma)^2 &=\gamma \circ \rho \circ \gamma \circ \gamma \circ \rho \circ \gamma\\ &=\gamma \circ \rho \circ \gamma \circ \rho \circ \gamma\\ &\subseteq \gamma \circ \gamma \circ \rho \circ \gamma \circ \gamma\\ &=\gamma \circ \rho \circ \gamma, \end{align} whence $(\gamma \circ \rho \circ \gamma)^n=\gamma \circ \rho \circ \gamma$, for all $n\geq1$, by induction, yielding $$\gamma \vee \rho = \bigcup_{n \geq 1} (\gamma \circ \rho)^n \subseteq \bigcup_{n\geq 1} (\gamma \circ \rho \circ \gamma)^n = \gamma \circ \rho \circ \gamma,$$ while the reverse inclusions is obvious.

I hope you can justify the several steps.
For example, I'm using the fact that composition of binary relations is associative.
In several ones I use $\gamma \circ \gamma = \gamma$, where one inclusion follows from transitivity, while the other follows from reflexivity.
In general, for two binary relations, $\alpha$ and $\beta$, if $\alpha$ is reflexive, then $\alpha \subseteq \alpha \circ \beta$.
I think these hints justify all the steps.
Let me know if you still have any doubts.