The congruence $\{(a^m, a^{m+r})\}^\#$ on $a^+$. I've spent a bit too long on this exercise. It's time to ask for help.
This is Exercise 1.20 of Howie's Fundamentals of Semigroup Theory.

Let $\rho_{m, r}$ (for $m, r\ge 1$) be the congruence $\{(a^m, a^{m+r})\}^\#$ on $a^+$. (Thus $a^+/\rho_{m, r}=M(m,r)$.)
(a) Show that $(a^p, a^q)\in\rho$ iff $p, q\ge m$ and $p\equiv q\pmod{r}$.

Here $\{(a^m, a^{m+r})\}^\#$ is the congruence generated by $\{(a^m, a^{m+r})\}$, $a^+$ is the free monogenic semigroup, and $M(m,r)$ is the monogenic semigroup of index $m$ and period $r$.
Let $\mathbb C=\{(a^{k+m+\ell}, a^{k+(m+r)+\ell})\mid k, \ell\in\mathbb{N}_0\}$. Then
$$\rho_{m, r}=\left[\mathbb{C}\cup\mathbb{C}^{-1}\cup 1_{a^+}\right]^{\infty}$$ is the equivalence generated by $\mathbb{C}$ on $a^+$.
Now if $(a^p, a^q)\in \rho=\rho_{m, r}$, there is some $n\in\mathbb{N}$ with $$(a^p, a^q)\in\left[\mathbb{C}\cup\mathbb{C}^{-1}\cup 1_{a^+}\right]^{n}$$ but then either I get what I'm after or $(a^p, a^q)\in 1_{a^+}$, i.e., $p=q$ is not necessarily greater than or equal to $m$. So perhaps we should assume $p\ne q$.
As for the converse, I've written down a lot but I haven't got much worth sharing. 

(b) Show that for all $m, n, r, s\ge 1$, $\rho_{m, r}\subseteq\rho_{n, s}$ iff $m\ge n$ and $s\mid r$.

I've tried a number of different approaches here that hinge on $$M(n, s)\simeq\frac{a^+}{\rho_{n, s}}``\,\subseteq\," \frac{a^+}{\rho_{m, r}}\simeq M(m, r),$$ which I'm not sure I'm allowed to assume (in the spirit of the question). These approaches are mostly pictorial and therefore difficult to describe, like this standard picture for $M(n, r)$:


(c) Deduce that, for all $m, n, r, s\ge 1$,
  $$\begin{align}\rho_{m, r}\cap\rho_{n, s}&=\rho_{\operatorname{max}(m, n), \operatorname{lcm}(r, s)}, \\
\rho_{m, r}\vee\rho_{n, s}&=\rho_{\operatorname{min}(m, n), \operatorname{hcf}(r, s)}.\end{align}$$

Here $\rho_{m, r}\vee\rho_{n, s}=(\rho_{m, r}\cup\rho_{n, s})^e$ is the equivalence generated by $\rho_{m, r}\cup\rho_{n, s}$.
I'm afraid, again, I don't have much (in fact anything) nontrivial to say despite my efforts. (I'm also running out of time allocated for this question.) I'm sorry.
Please help :)
 A: For (a), the result holds of course only for $p \not= q$.
For (b), it suffices to interpret $m$ and $r$ as semigroup parameters. Recall that the $\mathcal{J}$-depth of a monoid $M$, denoted $d(M)$, is the length of the longest $<_\mathcal{J}$ chain in $M$. The minimal ideal of a finite monoid $M$, denoted $I(M)$, is the unique minimal ideal for the $<_\mathcal{J}$ order. Now, $m = d(M(m,r))$ and $r$ is the size of $I(M(m,r))$, which in this case is isomorphic to the cyclic group $C_r$ of order $r$.
Note that $\rho_{m, r}\subseteq\rho_{n, s}$ if and only if $M(m,s)$ is a quotient of $M(n,r)$.
Moreover, if $N$ is a quotient of $M$, then $d(N) \leqslant d(M)$ and $I(N)$ is a quotient of $I(M)$. In particular, if $M(m,s)$ is a quotient of $M(n,r)$, then
$n = d(M(m,s)) \leqslant d(M(n,r)) = m$ and $C_s$ is a quotient of $C_r$, which implies that $s$ divides $r$. Conversely, if $n \leqslant m$ and $s$ divides $r$, I let you verify that $M(m,s)$ is a quotient of $M(n,r)$.
(c) Every congruence on $a^+$ is of the form $\rho_{m, r}$ for some $m \geqslant 0$ and $r > 0$. Let $\rho_{h, t} = \rho_{m, r}\cap\rho_{n, s}$. Then $\rho_{h, t}$ is the largest congruence contained in both $\rho_{m, r}$ and $\rho_{n, s}$. Thus by (b), $\rho_{h, t}$ is the largest congruence such that $h \geqslant m$,
$h \geqslant n$ and $t$ is a multiple of both $r$ and $s$. This means that $h = \max \{m, n\}$ and $t = \text{lcm}\{r, s\}$. The other case is similar.
