If $(\mathbb{Z}, +) \cong (C_\infty, \cdot)$, then $\exists K$ such that $K \trianglelefteq C_\infty$ and $K \cong 2\mathbb{Z}$?

Question

If $(\mathbb{Z}, +) \cong (C_\infty, \cdot)$, then does there exists a normal subgroup from $C_\infty$ (call it $K$) such that $K \cong 2\mathbb{Z}$?

I tried the following:

If $q \in C_\infty$, then $\langle q^2 \rangle := \{q^{2k} | k \in \mathbb{Z}\} \le C_\infty$.

Let $\beta: 2\mathbb{Z} \to \langle q^2 \rangle$ such that $\beta (m) = m^2$.

Unfortunately, this is not a homomorphism.

Important to note: This is just a curious question, NOT a homework.

Any hints will help.

EDIT

I think it'll be $\beta (2) = q^2$, where $q \in C_\infty$. Is that correct?