Consider $\mathbb{R}$ as a group under addition. Suppose that $G$ is a subgroup of $\mathbb{R}$ which is maximal w.r.t. the property that $1\not\in G$. Show that there is a unique prime number $p$ such that $p\in G$.
My attempt:
Suppose that there are two different prime numbers $p,p'\in G$. Then $1 = pm+p'n\in G$ (for some $m,n\in\mathbb{Z}$, Bézout).
So, if there is such a prime number, then it is necessarily unique. But how do I show that it exists? This question is given in a logic course (in fact, I had to prove the existence of the subgroup $G$ as above using Zorn's Lemma), so I don't think that I will have to use complicated algebra facts.
The subgroup $G$ is infinite, since, for example, $2\mathbb{Z}$ satisfies this property and $G$ is maximal (so $G\ge 2\mathbb{Z})$. Are $p\mathbb{Z}, p$ prime, the only subgroups satisfying this property? In that case, there always a unique prime $p$ in each $p\mathbb{Z}$, proving the statement.
Thanks.
Hint 1: If $G \cap \mathbb Z = \{ 0 \}$, define $G'= \langle G, 2 \rangle$ to be the subgroup of $G$ generated by $2$ and $G$. Show that $G'$ is a larger group which does not contain 1, contradicting the maximality.
Hint 2: You know that $H:= G \cap \mathbb Z \neq \{0\}$. Since $H$ is a subgroup of $\mathbb Z$ we have $H= n \mathbb Z$.
Show that if $p$ is a prime, $p\mid n$, then $G'=\langle G, p \rangle$ is also a subgroup of $\mathbb R$ which does not contain 1. Use maximality.