Does the set of integer powers of $\pi$ contain 0?

Question

The title, but more specifically, if I were to define the group $\langle \{\pi^n | n \in \mathbb{Z}\}, + \rangle$, would this satisfy the requirement for forming a group of containing an identity element? I am aware that this wouldn't form a group regardless since it isn't closed under addition.

Answer 1

No, it would not. For one, it does not contain $0$, as you contemplated. (Indeed, $\pi^n > 0$ for all $n \in \Bbb Z$.)
$+$ is not even a binary operation on the set. In fact, the sum of two elements of your set is never in the set. Indeed, if there existed integers $a, b, c$ such that $$\pi^a + \pi^b = \pi^c,$$ then multiplying with a large positive power of $\pi$ would yield $$\pi^A + \pi^B = \pi^C$$ for positive integers $A, B, C$. But $\pi$ is transcendental and so, it cannot satisfy the integer polynomial $x^A + x^B - x^C$.

Answer 2

Let $S = \{\pi^n \mid n \in {\Bbb Z}\}$. You consider the additive group $\langle S \rangle$ generated by $S$. This is not very precise, but I assume you mean the subgroup of $({\Bbb R}, +)$ generated by $S$. And in this case, yes, $\langle S \rangle$ contains the identity of the ambiant group, i.e. $0$.