a question on continuity of $G$-module for a profinite group $G$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:22:10Z http://mathoverflow.net/feeds/question/78469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78469/a-question-on-continuity-of-g-module-for-a-profinite-group-g a question on continuity of $G$-module for a profinite group $G$ unknown (google) 2011-10-18T16:47:15Z 2011-10-18T22:21:44Z <p>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?</p> http://mathoverflow.net/questions/78469/a-question-on-continuity-of-g-module-for-a-profinite-group-g/78470#78470 Answer by Wanderer for a question on continuity of $G$-module for a profinite group $G$ Wanderer 2011-10-18T17:10:38Z 2011-10-18T22:15:40Z <p>Not every finite index subgroup is open, but closed subgroups of finite index are open.</p> <p>So if the stabilizer is closed, that would be sufficient...</p>