Monoids in which $aN = Na$ and $ab \in N \leftrightarrow ba \in N$ aren't equivalent.

Question

Proposition. Let $G$ denote a group, and $N$ denote a subset of $G$. Then the following conditions are equivalent:

1. $aNa^{-1} = N$

2. $aN = Na$

3. $ab \in N \leftrightarrow ba \in N$

Proof.

$(1 \leftrightarrow 2).$ Obvious.

$(3 \rightarrow 2)$ Suppose $b \in aN$. Then $a^{-1}b \in N$. So $ba^{-1} \in N$. So $b \in Na$. The other direction is similar.

$(2 \rightarrow 3)$ Suppose $ab \in N$. Then $b \in a^{-1}N$. So $b \in Na^{-1}$. So $ba \in N$.

Now observe that $2$ and $3$ makes sense even if $G$ is just a monoid. They probably aren't equivalent at this level of generality, however.

Question. What are some example monoids $M$ for which $2$ and $3$ aren't equivalent?

Answer 1

Let $M = \{1, e, a, b, 0\}$ where $e^2 = e$, $aa = ab = 0$, $ae = b$, $ba = bb = 0$, $be = b$, $ea = eb = 0$, $1x = x1 = x$ and $0x = x0 = 0$ for all $x \in M$. Take $N = \{e\}$. Then $N$ is a submonoid of $M$ (it is even a subgroup) and one has $xy \in N \iff yx \in N$. However, $aN = \{ae\} = \{0\}$ but $Na = \{ea\}$.