Is the intersection of subsemigroups of a monoid $M$ that are also monoids, but with different identities from $M$, also a monoid?

Question

As an example, consider the associative operation $\ast$, where $$ a \ast b = ab \mod{30}. $$ Note that $\ast$ is closed and associative on the sets $$S_1 = \{0, 1, 2, \dots , 28 , 29\}$$ $$S_2 = \{0, 2, 4, \dots , 26, 28\}$$ $$S_3 = \{0, 3, 6, \dots ,24 , 27\}$$ $$S_4 = \{0, 6, 12, 18, 24\}$$ and, in addition, that $S_4 = S_2 \cap S_3$, so that $(S_4, \ast)$ is a subsemigroup of $(S_3, \ast)$ and so on. However, all of these monoids have different identities (in particular, $1, 16, 21,$ and $6$). Nonetheless, the intersection of the monoids $(S_2, \ast)$ and $(S_3, \ast)$ is also a monoid, although its identity is different. Will this always be the case?

Answer 1

No.

To see why, let $S$ be your favorite semigroup that isn't a monoid.

Now let's formally add elements $b$ and $c$ so that

• $bs = s = sb$ for each $s \in S$
• $cs = s = sc$ for each $s \in S$
• $b^2 = b$
• $c^2 = c$
• we can say nothing about how $b$ and $c$ multiply with each other (so they're added freely, except for the above conditions).

Then we have a semigroup $S'$ with two subsemigroups: $\langle S, b \rangle$ and $\langle S, c \rangle$, both of which are monoids, and whose intersection is $S$. All that's left is to ensure $S'$ is actually a monoid, which we can do by formally adjoining an element $1$ which acts as the identity everywhere.

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Concretely, let's look at the monoid

$$\langle a, b, c \mid ba = a = ab, \ ca = a = ac, \ b^2 = b, \ c^2 = c \rangle$$

of words in the alphabet $\{a,b,c\}$ (including the empty word $1$) subject to the above identities.

Then $\{ a^n \mid n \geq 1 \} \cup \{ b \}$ and $\{ a^n \mid n \geq 1 \} \cup \{ c \}$ are both subsemigroups which happen to be monoids. Yet their intersection is $\{ a^n \mid n \geq 1 \}$, which has no identity element.

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I hope this helps ^_^