why does $|S/G|=|S|-|G|+1$ imply that?

Because $\{\{x\} : x\in S\setminus G\}\cup \{G\}$ has size $|S|-|G|+1$

If there is an equivalence class one size larger than any of the above, then $|S/G|$ will be smaller

$G$ must be contained in one of the equivalence classes

um... not sure if that makes it clearer, hope it does

The cases I'm wondering about now are $|S| = 5, |G| = 2$ and ($|S/G| = 2$ and $|I| = 3$) or $|S/G| = 3$

Would be helpful to know how a semigroup with $3$ elements and no proper ideals looks like

For $|G| = 2$, case of $|S/G| = 2$ and $|I| = 1$ is like one above, case of $|S/G| = 2$ and $|I| = 2$ follows from if $S = \{1, a, b, c, d\}$, $a^2 = 1$ and $I = \{b, c\}$ since $b\sim c$ its clear we need $b\sim c\sim 1$ but $d\sim d\cdot b\in I$

So the latter is actually impossible

ah, I see, that's cute

is this bound achievable?

which one

oh

the one on $|S/G|$

oh wait, it's achieved for $|G|=1$...

and for $G=S$

No it isn't

I'm pretty sure not for monoids of order four

@robjohn looking forward to your "mid-March" apparel!

it seems like a suboptimal bound, but at the same time, monoids can behave very poorly

I mean, lol. I'm just need it to find one semigroup

It did good since it helped me to refute all semigroups of order four

ah no, I was missing the forest for the trees

Consider $f(x)=e^{i\alpha x}$ for $-\pi<x<\pi$ and $f(x+2\pi)=f(x)$, where $\alpha$ is not an integer. I'm supposed to study its Fourier series and verify $$\frac{\pi}{\sin \pi\alpha}=\frac{1}{\alpha}+\sum_{n=1}^\infty \frac{2(-1)^n\alpha}{\alpha^2-n^2},\qquad \left(\frac{\pi}{\sin \pi\alpha}\right)^2=\sum_{n=-\infty}^\infty\frac{1}{(\alpha-n)^2}.$$

I have verified the left formula, namely by setting $x=0$ in its Fourier series $$e^{i\alpha x}= \sum_{n\in\mathbb Z}\frac{\sin(\pi(\alpha-n))}{\pi(\alpha-n)}e^{inx}.$$

For the right-hand formula, what I've done is convolute $f$ with itself, which is just $f$ and then the coefficients become squared, but I must be doing something wrong because I'm not getting the desired result. Considering the convolution at $x=0$, we have $$1=\sum_{n\in\mathbb Z}\left(\frac{\sin(\pi(\alpha-n))}{\pi(\alpha-n)}\right)^2.$$ How can I get rid of that $n$ in the $\sin$?

take any group $G$ and any semigroup $G'$, then $G\coprod G'$ with multiplication induced by the multiplications of $G$ and $G'$ and letting every element of $G'$ absorb all of $G$ is a monoid $S$ with $s(S)=G$ and $S/G=\{G\}\cup\{\{x\}\colon x\in S\setminus G\}$, hence $|S/G|=|S|-|G|+1$

@Thorgott No!

I refuse to take any group!

I have STANDARDS!