Is $(\mathbb{N}, x^y)$ a monoid?

Question

Is $A = (\mathbb{N}, x^y)$ a monoid or a group?

I think the identity element in $A$ is $e = 1$. So it should be a monoid but I know it can't be a group as the inverse $a$ of is $0$. Is my understanding correct?

Answer 1

Let $x * y = x^y$. Then the structure $(\mathbb{N}, *)$ is not even a semigroup (that is, $*$ operation is not associative), and hence is not a monoid, since a monoid is a semigroup with the identity element. To prove it, note that $$(2 * 1) * 2 = (2^1) * 2 = 2 ^ 2 = 4 \neq 2 = 2^1 = 2 * (1 ^ 2) = 2 * (1 * 2).$$

Also, as noted by André Nicolas, there is no identity element $e$ for this operation. This also shows that $(\mathbb{N}, *)$ is not a monoid. Such an algebraic structure with only one binary operation is called a magma (or a groupoid).