In any topological group it is easy to show open subgroups are closed.

Prove that the converse holds in the following situation:

Let G be an abelian group, and let $$G=G_0 \supset G_1 \supset G_2\supset \dots$$ be a chain of subgroups. We turn $G$ into a topological group by defining the following topology: $U \subset G$ is open if and only if for every $x \in U$ there is some $n \in \mathbb{N}$ such that $x+G_n \subset U$.

I want to show that every closed subgroup of $G$ is open.

Possible useful facts:

  • The open subsets of $G$ are precisely unions of cosets of the $G_i$'s.
  • A subgroup $K \subset G$ is open if and only if there is some $n \in \mathbb{N}$ with $G_n \subset K$.

This is false.

Consider $G=\mathbb{Z}$, with $G_n=p^n\mathbb{Z}$. Then $\{0\}$ is a closed subgroup which is not open.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.