Existence of a commutative inverse semigroup with no identity element

Question

Does there exist a commutative inverse semigroup with no identity element, or are all commutative inverse semigroups abelian groups? If there does, what would be an example of such a commutative inverse semigroup? If there does not, then how does one prove that all commutative inverse semigroups are abelian groups?

Answer 1

The set of natural numbers $\mathbb{N}$ equipped with the binary operation $x \cdot y = \min(x, y)$, where $\min(x, y)$ is the minimum of $x$ and $y$, is a commutative inverse semigroup with no identity element. By definition, $\min(x, y)$ is closed over $\mathbb{N}$, commutative, and associative. As each natural number is idempotent under $\min(x, y)$, each natural number satisfies $x \cdot x \cdot x = x$ and is its own inverse according to the definiton of an inverse in an inverse semigroup, as two elements $x$ and $y$ are inverse elements of an inverse semigroup if $x \cdot y \cdot x = x$ and $y \cdot x \cdot y = y$, and if $x = y$, then $x \cdot x \cdot x = x$. $\min(x, y)$ does not have an identity element, as every natural number $n$ is an absorbing element when $\min(x, y)$ is restricted to the subset $\mathbb{N} + n$. This makes the heretofore mentioned set a commutative inverse semigroup with no identity element.

Answer 2

Minimal example. Let $S = \{a, b, 0\}$ with $a^2 = a$, $b^2 = b$ and all other products equal to $0$. Then $S$ is inverse and commutative but is not a monoid.