a question on continuity of $G$-module for a profinite group $G$

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I have seen the following statment somewhere, for example in Appendix B2 on Silverman's book "The Arithmetic of Elliptic Curves" : Let $M$ be an abelian group with discrete topology and $G$ be a profinite group. Then an linear action ( which means that $\sigma(m_1+m_2)=\sigma(m_1)+\sigma(m_2)$, i.e it is a $G$-module) $\phi : G \times M \rightarrow M$ is continuous if and only if the stabilizer $\sigma \in G | \sigma(m)=m$ has finite index in $G$ for all $m \in M$. But what we need is that this stabilizer is open in $G$. I also saw that in a profinite group, not every subgroup of finite index is open. So is this statement correct? Or how to see that this stabilizer is open if it has finite rank?

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If the group is finitely generated, finite-index subgroups are open:ams.org/mathscinet-getitem?mr=2276769 – Agol Oct 18 2011 at 17:13

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It is false as stated. For instance, consider any faithful action of a quotient of G by a non-closed finite index subgroup. "Finite index" should be replaced with either "open" or "closed and finite index". – Kevin Ventullo Oct 18 2011 at 21:47
@Kevin Ventullo: Thanks. But I made a mistake in my question. In fact, the statement is for $G$-modules, i.e it requires the action satisfies $\sigma (m_1 + m_2) = \sigma(m_1) + \sigma(m_2)$. The example that $G$ acts on $G/H$ for a subgroup $H$ of $G$ is not a $G$-module. I should add this condition in my question, sorry. – unknown (google) Oct 18 2011 at 22:18

1 Answer

Not every finite index subgroup is open, but closed subgroups of finite index are open.

So if the stabilizer is closed, that would be sufficient...

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Without continuity, I can't see why the stabilizer is closed....:( Could you give me a hint? Thanks. – unknown (google) Oct 18 2011 at 19:05
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