Maximal implies prime in an integral po-groupoid [is false!]

Question

In "Lattice Theory" by G. Birkhoff, in XIV.9 there is the following Lemma 1:

In any integral po-groupoid, any maximal element is prime and indecomposable

A po-groupoid is a poset equipped with a binary operation $\cdot$ such that $a\leq b$ implies $xa \leq xb$ and $ax\leq bx$. It is integral if $1\geq a$ for all $a$. A maximal element is then an element covered by $1$ and $p < 1$ is prime if $xy\leq p$ implies $x\leq p$ or $y\leq p$.

The proof of the first part goes as such:

[...] Again, unless $x\leq m, x\vee m = 1$. Hence if $xy\leq m$ but $x\nleq m$, then $y = 1y = (x\vee m)y \leq_! xy \vee my = m \vee m1 = m$ so that $m$ is [a] prime.

I don't understand where the inequality marked with $_!$ comes from. My understanding is that the reverse inequality is true, but not this one.

Am I missing something obvious here?

Answer 1

I think the theorem is false.

For a counterexample, consider the $4$-element set $$G = \{0,a,b,1\}$$ with multiplication defined by

• $\,1x = x1 = x$, for all $x$.$\\[4pt]$
• $\,xy=0\;$if neither of $x,y\;$is $1$.

and partial order defined by

• $\,0 \le x \le 1$, for all $x$.$\\[4pt]$
• $\,a,b\;$not comparable.

Then $a,b\;$are both maximal, but neither is prime.

For example, letting

$\qquad x = a,\;y = a,\;m = b$

we have

$\qquad xy \le m$, but $x \not\le m,\;y \not\le m$.