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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.

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    do you mean $1\notin G$ at the start? – 2017-02-16
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    Excuse me! I've fixed the mistakes. I do indeed mean $(\mathbb{R},+)$. $\mathbb{R^+}$ was the notation used by my Group theory course book, but I realize that $(\mathbb{R},+)$ is probably less ambiguous. – 2017-02-16

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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.

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    In that sense, given any chain consisting of $H_i$, it's possible to prove that: $\cup_{i \in I} H_i$ itself must be a group. Does that suffice to conclude that all chains have an upper bound? Then Zorn's lemma answers the question. – 2017-02-17
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    @Edwin: Yes -- except you still need to prove that $\cup H_i$ is a subgroup _of the particular form you're looking for_ -- which is of course very easy in the case of "does not contain $1$". – 2017-02-17
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    What should be the quotient $\mathbb{R}/G$? – 2018-01-20
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    @Janson: I _think_ it will always be isomorphic to $\mathbb Z[\frac 1p]/\mathbb Z$ for some prime $p$, but I'm not completely sure of all steps in my reasoning. – 2018-01-20
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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

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    A chain does not need to be nicely indexed by $\mathbb N$ like you present it here -- in order to apply Zorn's lemma you need to prove upper bounds for _any_ totally ordered subset of the partial order. – 2017-02-16
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    You're quite right. I did have the union in mind and mistakenly wrote it as an indexed sequence. Thank you for the correction – 2017-02-17