Does this proof contain a contradiction?

Question

I have made the following proof and I am asking if there is anything wrong in my steps:

Let $f:\mathbb{R}\to\mathbb{R}$ be a real analytic function with infinitely many zeros. Let $a\neq 0$ be a real number. We know that the set $f^{-1}(a)$ is discrete. I want to make a bijection between (not necessarily an isomorphism) $f^{-1}(a)$ and a finite Abelian group $T$. I have two cases:

1) If the set $f^{-1}(a)$ is finite, then a bijection exists if the two sets $f^{-1}(a)$ and $T$ have the same number of elements.

2) If the set $f^{-1}(a)$ is infinite, then we can find a bijection between $f^{-1}(a)$ and $ℤ$, hence we can mod out (http://en.wikipedia.org/wiki/Modulo_(jargon)) $ℤ$ by an integer $n≠0$ to obtain a smaller and finite structure whose elements are the classes, namely we obtain the finite group $ℤ/nℤ$. Hence in this cases we can construct a bijection between $ℤ/nℤ$ and a finite Abelian group $T$.

Answer 1

I have to object to the second part of your proof. If you have an infinite set, you can't construct a bijection with a finite set, essentially by definition. It also seems to me that it woud be more natural in this problem to consider a bijection between $f^{-1}(a)$ and some finite set $S$, as you are not trying to construct an isomorphism (or homomorphism). The end result would be the same, and it seems that this would be less confused. Trying to relate the set $f^{-1}(a)$ to a group is only causing confusion: the part when you construct a bijection and then mod out doesn't make sense (in the context of this problem/proof). If you want to show that there exists a bijection between $f^{-1}(a)$ and a finite set, you will have to show that case $2$ cannot occur.