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In this answer, a subgroup $H$ is considered to be a assumed to be of finite index $n$ in $\Bbb Q$. The author makes two claims:

  • $n\Bbb Q = \Bbb Q$
  • $nH=H$ (since $n(q+H) = H$ where $q\in \Bbb Q$)

I'm not sure why $n\Bbb Q = \Bbb Q$ and $nH = H$.

Or for that matter why is $n(q + H) = H$ or $nq + H = H$? I mean, I'm not sure we can say $nq = 0$ unless we know for certain that $\Bbb Q/H$ is cyclic of order $n$. Do we know that $\Bbb Q/H$ is cyclic?

Could someone please explain?

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    Those are not the claims they make. They say $n\Bbb Q=\Bbb Q$ and $n\Bbb Q\subseteq H$. – Arthur May 02 '20 at 16:04
  • @Arthur Surely they say that $n(q + H) = H \implies nq + nH = H$ though, which implies $nH = H$, since $nq + H = H$? –  May 02 '20 at 16:06
  • Which is the element in the quotient group $\Bbb Q/H$ in this context? $q$? –  May 02 '20 at 16:14

2 Answers2

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Take an arbitrary rational number $\frac ab$. Then $$\frac ab=n\cdot \frac a{nb}\in n\Bbb Q$$ This shows that $n\Bbb Q=\Bbb Q$.

Next they claim that $n\Bbb Q\subseteq H$. They show this by pointing out that in the quotient group $\Bbb Q/H$, multiplying any arbitrary element $q+H$ by $n$ gives you the identity element $H$, by Lagrange's theorem. This means we must have $qn\in H$.

Arthur
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  • Thanks! How exactly does it follow from Lagrange's theorem that multiplying any arbitrary element $q+H$ in the quotient group $\Bbb Q/H$ by $n$ gives the identity element $H$? It's not really obvious to me. All Lagrange's theorem says is that "the order (number of elements) of every subgroup H of G divides the order of G". –  May 02 '20 at 16:20
  • @Triskelion The quotient group has order $n$. Thus Lagrange's theorem says that the order of any element is divisible by $n$ (it says a little more than that, but that's all we need). – Arthur May 02 '20 at 16:23
  • So the order of the element $q + H$ in $\Bbb Q/H$ is some divisor of $n$. OK. Then $n(q+H) = H$. But how does it follow from $n(q+H)=H$ that $qn \in H$? –  May 02 '20 at 16:29
  • @Triskelion Because $qn+H$ and $H$ are equal as elements of the quotient group. That's what equality means there. – Arthur May 02 '20 at 16:31
  • Well, I precisely don't understand that part. How does it follow from $n(q+H)=H$ that $nq + H = H$? –  May 02 '20 at 16:32
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    Oh, I see. $H + H = H$ due to closure property of groups. So $(q + H) + (q + H) = 2q + H$ and $(q + H) \times n = nq + H$. –  May 02 '20 at 16:37
  • @Triskelion Yup. That's what the group operation looks like in a quotient group (of an additive group). ($H$ is really $0+H$.) – Arthur May 02 '20 at 16:42
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See $ \mathbb Q $ is a divisible group hence $n \mathbb Q= \mathbb Q$ And the second claim there was $n\mathbb Q \subseteq H$. It follows from the hypotheses and Lagrange's theorem.

Naba Kumar Bhattacharya
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  • I don't know what divisible group means. It would be better if you could explain from first principles. –  May 02 '20 at 16:12
  • A divisible group is a nontrivial abelian group G such that for every $n>0$ integer , $G \rightarrow G$ multiplication by n map is surjective. Prove this for $\mathbb Q $ and hence prove that for any divisible group $G$, $nG=G$ for all $n>0$. – Naba Kumar Bhattacharya May 02 '20 at 16:16