Prove, for ever $x$ in a maximal subgroup there is a $n$ such that $p^n x \in G$. The third part of this question is giving me problems. I think I'm missing something trivial (?).

Consider $\mathbb{R}$ as a group under addition.

*

*Prove, using Zorn's Lemma, that there is a subgroup $G$ of $\mathbb{R}$ which is maximal w.r.t. the property that $1\not \in G$.


*Suppose $G$ is as in (1.). Show that there is a unique prime number $p$ such that $p\in G$.


*Let $p$ be as in (2.). Prove that for every $x \in \mathbb{R}$ there is an $n\geqslant 0$ such that $p^n x\in G$.

Proof
I've been able to prove 1. and 2. but for 3. I seem to be missing a way of putting 1. and 2. together. I was thinking of the following:
Let $G$ be a maximal subgroup with $1\not \in G$. Choose an $x \in \mathbb{R}$. If $x\in G$ then $n=0$ is enough.
Now let $x\not \in G$. Consider $H:= \langle G \cup \{x\}\rangle$. Then $1\in H$ because $G$ was maximal. Which implies $1=g+k\cdot x$ for a certain $g\in G$ and $k\in \mathbb{Z}$.
This implies $p = p\cdot g + pkx$ which implies $pkx \in G$ (because $p\in G$).
However, how should I continue, how can I use (2.)? (the fact that $p$ is the only prime in $G$)? Any tips?
EDIT
Inspired by Derek Holt, I completed the proof as follows:
First notice that 2. implies $G\cap \mathbb{Z} = \langle p \rangle$.
Now once again let $x\not \in G$ and consider $\langle G \cup \{x\}\rangle$, then $1$ must be in this group, which implies $1 = g_0+k_0x$ for certain $g_0, k_0$. Or $p=p\cdot g_0+pk_0x$, which implies $pk_0x \in G$.
Now consider $px$, if $px\in G$ we are done, else consider $\langle G \cup \{px\}\rangle$ which must contain 1 due to maximality of $G$. So $1 = g_1+\tilde k_0px$ for certain $g_1, \tilde k_0$. Which implies $\tilde k_0 px -1 \in G\Rightarrow k_0 \tilde k_0px - k_0 \in G$ but since $k_0px\in G$ combining these results in $k_0 \in G$. Howver $k_0 \in \mathbb{Z}$. Thus $k_0 \in G\cap \mathbb{Z} = \langle p\rangle$. I.e. $k_0 = k_1\cdot p$ for a certain $k_1$ where $|k_1| < |k_0|$.
This implies $k_0px = k_1p^2x \in G$.
This process can be continued untill $|k_{n-1}| = 1$ which results in $p^nx \in G$.
 A: If $x \in G$ we are done so suppose not. Then by maximality of $G$, we must have $1 \in G + \langle x \rangle$, so there exists $k_0 \in {\mathbb Z}$ with $k_0 x - 1 \in G$, and hence $k_0 p x \in G$.
Similarly, if $px \in G$ we are done so suppose not. Then $1 \in G + \langle px \rangle$, so there exists $k_1 \in {\mathbb Z}$ with $k_1 p x - 1 \in G$. Hence $(k_1 + rk_0)px - 1 \in G$ for all $r \in {\mathbb Z}$, and so we can choose $k_1$ with $|k_1| < |k_0|$. Then $k_1 p^2 x \in G$.
Carrying on like this, if $p^2x \not\in G$ then we find $k_2$ with $|k_2| < |k_1|$ and $k_2p^2x - 1 \in G$, etc, and since $|k_0| > |k_1| > |k_2| > \cdots$, the process must eventually stop with $p^nx \in G$ for some $n$.
A: Here is an alternative proof of 3, which I like better. Let $Q = {\mathbb R}/G$ be the quotient group. Then the image $\bar{1}$ of $1$ in $Q$ has order $p$. The maximality condition implies that $\langle \bar{1} \rangle$ is contained in all nontrivial subgroups of $Q$.
Part 3 says that all elements of $Q$ have order a power of $p$. Suppose not. Then there exists $x \in {\mathbb R}$ such that the order of its image $\bar{x}$ in $Q$ is either infinite or is a prime other than $p$. In either case $\langle \bar{x} \rangle$ has no subgroup of order $p$, so cannot contain $\langle \bar{1} \rangle$, contradiction.
