Is $(\Bbb N,\operatorname{LCM}(a,b))$ a monoid?

Question

Is $(\Bbb N,\operatorname{LCM}(a,b))$ a monoid?

Is it associative and does it have a neutral element?

Refer to answer below for my attempt.

Answer 1

The neutral element cannot depend on elements: it is a global notion.

Moreover your problem is ill-posed: $\operatorname{lcm}(a,b)$ is not $0$, though $0$ is a common multiple of all numbers, so you question should be : Is $(\mathbf N^*, \operatorname{lcm})$ a monoid? The answer is yes: as you showed, it is associative, and the neutral element is $1$.

Note that it is not a regular monoid: $\operatorname{lcm}(16,6)=\operatorname{lcm}(16,12)$.

Answer 2

The neutral element of each element is itself, since $\operatorname{lcm}(a,a)=a$ and also $\operatorname{lcm}(\operatorname{lcm}(a,b),c)=\operatorname{lcm}(a,\operatorname{lcm}(b,c))$, hence it is associative, thus $(\Bbb N, \operatorname{lcm}(a,b))$ is a monoid.

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$\operatorname{lcm}(\operatorname{lcm}(a,b),c)=\operatorname{lcm}(a,\operatorname{lcm}(b,c))$

Proof

Answer 3

In fact, we can add GCD and get a lattice.

The set $\mathbb N := \{0,1,2,3,\dots\}$ with the "divisibility" partial order $$ a \; | \; b \Longleftrightarrow \exists k \big(ak=b\big) $$ is a lattice. Lattice operations $a \vee b$ and $a \wedge b$ are LCM and GCD, respectively. The operations are associative. And (as I said in a comment), $0$ is the greatest element, while $1$ is the least element. So $0$ is the neutral element for $\wedge$; and $1$ is the neutral element for $\vee$.