Showing that $\mathbb{Q}$ has no subgroups of finite index, without using quotient groups. [duplicate]

So I've seen many solutions to this problem using the idea of a quotient group. In my class, however, we have no covered quotient subgroups, so I am looking for a different way to prove this. So far, my line of thinking has been to assume such a subgroup ($H$) exists, choose an arbitrary $q \in \mathbb{Q} \setminus H$, and show that we must have $q \in H$. If $H$ is of finite index, there must be finitely many cosets, so I started looking at some cosets:

$0+H \rightarrow$ This is just $H$, since $H$ is a group, so closed under operation.

$q+H \rightarrow$ This coset contains $q$, but cannot contain $2q$ or $0$, since that would imply that $q \in H$.

$2q+H \rightarrow$ This contains $2q$, but not $3q$ or $q$, since $q \notin H$.

$nq+H \rightarrow$ Contains $nq$, but not $(n-1)q$ or $(n+1)q$.

I've also noticed that if $H$ is to have finite index, then eventually the cosets of integer multiples of $q$ must repeat. That is

$$\exists\ k,m \in \mathbb{Z}\ \text{with}\ kq+H = mq+H$$

$$\iff$$

$$kq+h_1 = mq+h_2\ \text{for some}\ h_1,h_2 \in H$$ $$\iff$$ $$(k-m)q = h_2-h_1 \in H$$ $$\iff$$ $$\text{Some integer multiple of}\ q\ \text{is in}\ H$$

But I don't think this is a contradiction, so I'm not sure how to proceed.

I think that if I could show that every $q$ has an integer multiple, then we could show that every $q$ must be in $H$, since we would be able to express any rational number as an integer multiple of another one. However, this line of reasoning would only hold if the integer multiple was the same integer for any rational we choose (perhaps it would be the order of the subgroup?)

An answer or a hint would be hugely appreciated.

marked as duplicate by Dietrich Burde, user133281, Leucippus, Ali Caglayan, Parcly TaxelOct 24 '16 at 8:56

• What properties of cosets are you allowed to use? – rogerl Oct 23 '16 at 19:26
• What you do is the same as using quotient groups, because you are dealing with cosets forming a quotient group. I find the title strange. – Dietrich Burde Oct 23 '16 at 19:34
• @rogerl We can really only use the fact that the index of a subgroup refers to the number of cosets. We have learned many characteristics of finite groups (Lagrange's Theorem, for example), but I don't think they are relevant here. We know that cosets form a partition of the original group, and that all cosets are the same size. – Alex Oct 23 '16 at 19:48
• @DietrichBurde We have not learned what a quotient group is yet. If there is a reasonable answer without using the words "quotient group", then I suppose it would be acceptable. – Alex Oct 23 '16 at 19:49
• The quotient group $\mathbb{Q}/H$ is the set of cosets $\{q+H\mid q\in \mathbb{Q}\}$. So you can take the "so many solutions" which you have seen, and just not using the word "quotient group". – Dietrich Burde Oct 23 '16 at 19:58

Clearly the set $\frac{1}{p^n}$ such that $p$ is prime and $n>0$ generates $\mathbb Q$.
So a proper subgroup of $H$ cannot contain that set. Therefore there exists a prime $p$ an an integer $n$ such that $\frac{1}{p^n}$ does not belong to $H$.
Given $m>n$ consider the fractions of the form $\frac{a}{p^m}$. Suppose that $H\frac{a}{p^m}=H\frac{b}{p^m}$ then $\frac{a-b}{p^m}\in H$. Since $\frac{1}{p^n}\not \in H$ we conclude that $p^{m-n}| a-b$. Hence, if we consider $x\neq y\in \{1,2,3,\dots p^{m-n}\}$ we have that $H\frac{x}{p^m}\neq H\frac{y}{p^m}$.
Conclusion: $H$ has at least $p^{n-m}$ cosets for every $m>n$. This clearly implies $H$ has an infinite number of cosets.