Subsets of the integers which are closed under multiplication Let $S$ be a subset of the integers which is closed under multiplication. There are many possible choices of $S$:

*

*$S = \{-1, 1\}$.

*$S$ is the set of integers of the form $a^k$, where $a$ is fixed and $k \geq 0$ varies. For example, $S$ is the set of powers of 2.

*$S$ is the set of integers of the form $a^k$, where $a$ is varied and $k \geq 0$ is fixed. For example, $S$ is the set of all squares.

*$S$ is the set of integers divisible by some set of primes and not divisible by some other set of primes. For example, $S$ is the set of $B$-smooth integers, i.e. integers not divisible by any primes larger than $B$, or $S$ is the set of multiples of 13, or $S$ is the set of even numbers.

*Here is a more exotic choice: $S$ is the set of all integers which can be written as a sum of two squares (number theorists will recognize this as a special case of #4).

I have two questions: first, is there a simple reason why the collection of subsets of the integers which is closed under multiplication is so much richer than the collection of subsets of the integers which is closed under addition? Second, are there any other interesting choices of $S$ which aren't on my list?
 A: That is  because the semigroup $({\mathbb Z},\times)$ contains the semigroup $({\mathbb N},+)^\infty$ as an isomorphic copy. In contrast, most of the subsemigroups of $({\mathbb Z},+)$ are isomorphic to subsemigroups of $({\mathbb N},+)$.
A: To make Mark Sapir's answer more explicit, we can take the direct product of sets obtained from your second example. For instance, we can take $S = \{ x: \exists a \in A, b \in B: x =ab\}$ where $A = \{x:\exists k:  x=3^k\}$ and $B = \{x:\exists k:  x=5^k\}$.
If we consider just the cases where each component set is built around a power of a prime, and we only have one component set for each prime, then we can represent what power is associated with each prime by some function $f$ from the prime numbers to the non-negative integers, and we can define $S_p=\{x: \exists k : x = p^{kf(p)}\}$. This means that for every such $f$, there is some set constructed from the direct product of all $S_p$, so this gives a family of cardinality $\mathbb N ^ {\mathbb N}$.
If we take $f(p)$ to be a constant function $f(p) \equiv c$, then we get your third example of numbers of a fixed power. I believe that some of your other examples are outside this family, but as far as the cardinality is concerned, adding those examples doesn't change the number of sets.
