Prove $(\mathbb{R},+)$ without 1 has a maximal subgroup using Zorns lemma 
Prove, using Zorn’s Lemma, that there is a subgroup $G$ of $\mathbb{R}$ which is maximal w.r.t. the property that $1 \not\in G$. Assuming that $\mathbb{R}$ is a group under addition. 

My attempt thus far:
A subgroup is also obviously a subset of $\mathbb{R}$, as such we can use the powerset. Which is partially ordered by inclusion. 
Let $H \subset\mathcal{P}(\mathbb{R})$ be the set of all subsets which form a group under addition, but do not contain the element $1$. 
This group is obviously non-empty, due to the fact that $\langle 2^n\rangle$ forms such a subgroup for any $n \in \mathbb{N}$. So does $\langle \pi\rangle$. If I can prove that all chains in $H$ have an upper bound, Zorn's Lemma gives the desired result.  
I am however unable to prove that all chains have an upper bound, because I can't find a way to classify all chains. Obviously any irrational number spans a chain, by $y \in \mathbb{R \setminus Q}$ and $\langle ky\rangle$ for any $k>0$ which has a maximum element $\langle y\rangle$, similarly for every even number. I'm unable to classify the rational numbers because $\frac{1}{3} \in \mathbb{Q^+}$, $\frac{1}{3}^n=3*\frac{1}{3}=1$.
However, in this way I can't prove that every chain is found nor that they all have upper bounds.     
 A: Don't bother to try to classify chains or subgroups. Zorn's lemma doesn't demand that you know what they are -- instead the lemma is asking you to be prepared to be given a random chain, and then you must argue that it has an upper bound somewhere.
For this, you can simply note that whenever you have a chain of subgroups, its union will itself be a subgroup of the desired form. You can prove this without knowing how the subgroups look -- it is enough to know that they are subgroups and work directly from the definitions.
A: Note: it has been pointed out that this answer is not entirely correct. I'll leave it for prosperity, see Henning's answer for what I should have written
You don't need to classify all chains.
Suppose you have some chain $H_1<H_2<H_3<\cdots$ then $H=\lim\limits_{n\to\infty}H_n=\{h\in H_i|i=1,2,\ldots\}$ is the required upper bound. To check this is a group notice that any $x,y\in H$ must both be in some $H_i$ so $xy^{-1}\in H_i<H$. It cannot contain $1$ as $1\notin H_i$ for each $i$
