I was struggling with that last line of this proof. (this was found here: http://www-users.math.umn.edu/~felix077/download/sec3.pdf)
I understand that ker$(\varphi \sigma) =\varphi^{-1} (\sigma^{-1} (N)) $, and that this is a proper subgroup of $\mathbb{Q}$, but I don't understand how they can talk about its index in $\mathbb{Q}$.
The earlier problem said there are no proper subgroups of $\mathbb{Q}$ with a finite index, but aren't we dealing with $\mathbb{Q} / \mathbb{Z}$ here? If a subgroup has finite index in $\mathbb{Q} / \mathbb{Z}$ , does it have finite index in $\mathbb{Q}$?
Best Answer
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 $\mathbb{Q}/\mathbb{Z}$.