# homomorphism of profinite groups

Let $G$ be a profinite group and consider a continuous surjective homomorphism: $$\phi:G\rightarrow \widehat{\mathbb Z}$$ where $\widehat{\mathbb Z}:=\varprojlim \mathbb Z/n\mathbb Z$. Moreover Let $H<G$ be a (closed) subgroups such that $f=|\widehat{\mathbb Z}:\phi(H)|<\infty$. I Have two questions:

1. Why the map $\frac{1}{f}\phi:H\rightarrow\widehat{\mathbb Z}$ is well defined? I mean: why if $a\in\phi(H)$ then $\frac{1}{f}a\in \widehat{\mathbb Z}$? $f$ is a natural number, so its inverse is not supposed to be in $\widehat {\mathbb Z}$.
2. Why $\frac{1}{f}\phi:H\rightarrow\widehat{\mathbb Z}$ is surjective?

## 1 Answer

If $\phi(H)$ has index $f$ in $\widehat{\mathbb{Z}}$, then $\phi(H)$ must in fact be the subgroup $f \widehat{\mathbb{Z}}$, since it is the unique subgroup with that index. This answers both of your questions.

• How can I prove that the open subgroups (=of finite index) of $\widehat{\mathbb Z}$ are of the type $m\widehat{\mathbb Z}$ – Dubious Feb 6 '16 at 18:39
• @Dubious: it's equivalent to show that the only (continuous) finite quotients of $\widehat{\mathbb{Z}}$ are the quotients $\widehat{\mathbb{Z}}/f\widehat{\mathbb{Z}}$, which follows from the observation that they must be cyclic since $\widehat{\mathbb{Z}}$ is (topologically) cyclic. – Qiaochu Yuan Feb 6 '16 at 18:44