Consider the group $\mathbb{Q}$ under addition of rational numbers. If $H$ is a subgroup of $\mathbb{Q}$ with finite index, then $H = \mathbb{Q}$.
I just saw this on our exam earlier and was stumped on how to show this. Any ideas?
[Math] Subgroup of $\mathbb{Q}$ with finite index
abstract-algebragroup-theory
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Thanks to the hint of user1952009 I managed to solve the exercise with proof. Let $$X := \{ xH : x \in G\}$$ and consider the group action $$\begin{cases} G \times X\to X\\ (g,xH) \mapsto gxH\end{cases}$$ Naturally we can associate with each group action a bijection $$\lambda_g:\begin{cases}X\to X\\ xH \mapsto gxH\end{cases}$$ for $g \in G$ and a homomorphism $$\lambda:\begin{cases}G\to S_X\\ g \mapsto \lambda_g\end{cases}$$ where $S_x$ denotes the symmetric group of $X$. Now we show $$\ker \lambda = \bigcap_{g \in G}gHg^{-1}$$ Let $y \in \bigcap_{g \in G}gHg^{-1}$. Thus we can write $y = xh_x x^{-1}$ with $x \in G$ and $h_x \in H$. Therefore for any $x \in G$ we have $$\lambda_y(xH) = yxH = xh_xx^{-1}xH = xh_xH = xH$$ Conversly, if $y \in \ker \lambda$ we have $\lambda_y = \operatorname{id}_X$. Hence $$yxH = xH$$ for all $xH \in X$. Therefore $$yxh_1 = xh_2 \quad \Leftrightarrow \quad y = xh_2h_1^{-1}x^{-1} \in xHx^{-1}$$ for all $x \in G$. Thus $y \in\bigcap_{g \in G}gHg^{-1}$. By the isomorphism theorem we have $$G/\ker \lambda = G/\bigcap_{g \in G}gHg^{-1} \cong \operatorname{Im} \lambda \leq S_X$$ And since $[G:H]$ is finite, $X$ is finite of cardinality $[G:H]!$. Thus we conclude that $[G:\bigcap_{g \in G}gHg^{-1}]$ is also finite.
First of all, it is $\mathbb{Q}/\mathbb{Z}$, not $\mathbb{Q}\setminus\mathbb{Z}$. This is a quotient group.
Now, since $N\leq \mathbb{Q}/\mathbb{Z}$ has finite index then the quotient $(\mathbb{Q}/\mathbb{Z})/N$ is a finite group. $\varphi\circ\sigma:\mathbb{Q}\to(\mathbb{Q}/\mathbb{Z})/N$ is a homomorphism of groups. Hence by the first isomorphism theorem we have $\mathbb{Q}/Ker(\varphi\circ\sigma)\cong Im(\varphi\circ\sigma)\leq(\mathbb{Q}/\mathbb{Z})/N$. It follows that $\mathbb{Q}/Ker(\varphi\circ\sigma)$ is a finite group, so $Ker(\varphi\circ\sigma)$ has finite index in $\mathbb{Q}$. Now, why is it a proper subgroup? Because if the kernel would be all $\mathbb{Q}$ then we would get that the kernel of $\varphi$ (which is $N$) is all $\mathbb{Q}/\mathbb{Z}$, which is a contradiction to $N$ being a proper subgroup of $\mathbb{Q}/\mathbb{Z}$.
Best Answer
Show that if $[\Bbb Q:H]=n$, $nq\in H$ for every $q\in\Bbb Q$. Conclude that $n\Bbb Q\subseteq H$. But $n\Bbb Q=\Bbb Q$, so $H=\Bbb Q$.