Must a subgroup neighborhood of 0 in an abelian topological group be open?

Question

Here, it is written that if a topological abelian group $M$ has a fundamental system of neighborhoods of 0 consisting of subgroups, then the subgroups must be open.

Why is this true?

Answer 1

Alright, so the answer is yes:

Let $N\subset M$ be a subgroup which is a neighborhood of $0$. Then, there is an open $U\subset N$ containing $0$. Then, we have trivially $$N = \bigcup_{n\in N}(n+U)$$ where $n+U$ is the set $\{n+u : u\in U\}$, which is open due to continuity of the group operation. Thus, $N$ is a union of opens, and hence open.