Are normal subgroups of a profinite group with finite index closed?












1












$begingroup$


Let $G$ be a profinite group and $H subseteq G$ be a normal subgroup with $[G:H] < infty$, i.e. $H$ has finite index.



Question: Is $H$ closed?



Problems:



I have a lot of trouble to understand these profinite groups in general. I know that they are inverse limits where every finite group is endowed with the discrete topology. Then we say that this inverse limit is endowed with the product topology of these discrete groups which I find hard already since I only worked with finite products before. I really have no clue how open/closed sets look like.



The only proof I understood is that in a compact topological space, every open subgroup is exactly the closed subgroups of finite index. I heard that there is some relation to the normal subgroups, but I did not really understand it. I need something to understand the proof in this post.



Could you please explain this concept to me? Thank you in advance!






abstract-algebra general-topology group-theory topological-groups normal-subgroups







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$endgroup$












  • $begingroup$
    Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
    $endgroup$
    – Yanko
    Jan 6 at 13:04


















1












$begingroup$


Let $G$ be a profinite group and $H subseteq G$ be a normal subgroup with $[G:H] < infty$, i.e. $H$ has finite index.



Question: Is $H$ closed?



Problems:



I have a lot of trouble to understand these profinite groups in general. I know that they are inverse limits where every finite group is endowed with the discrete topology. Then we say that this inverse limit is endowed with the product topology of these discrete groups which I find hard already since I only worked with finite products before. I really have no clue how open/closed sets look like.



The only proof I understood is that in a compact topological space, every open subgroup is exactly the closed subgroups of finite index. I heard that there is some relation to the normal subgroups, but I did not really understand it. I need something to understand the proof in this post.



Could you please explain this concept to me? Thank you in advance!






abstract-algebra general-topology group-theory topological-groups normal-subgroups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
    $endgroup$
    – Yanko
    Jan 6 at 13:04
















1












1








1





$begingroup$


Let $G$ be a profinite group and $H subseteq G$ be a normal subgroup with $[G:H] < infty$, i.e. $H$ has finite index.



Question: Is $H$ closed?



Problems:



I have a lot of trouble to understand these profinite groups in general. I know that they are inverse limits where every finite group is endowed with the discrete topology. Then we say that this inverse limit is endowed with the product topology of these discrete groups which I find hard already since I only worked with finite products before. I really have no clue how open/closed sets look like.



The only proof I understood is that in a compact topological space, every open subgroup is exactly the closed subgroups of finite index. I heard that there is some relation to the normal subgroups, but I did not really understand it. I need something to understand the proof in this post.



Could you please explain this concept to me? Thank you in advance!






abstract-algebra general-topology group-theory topological-groups normal-subgroups







share|cite|improve this question











$endgroup$




Let $G$ be a profinite group and $H subseteq G$ be a normal subgroup with $[G:H] < infty$, i.e. $H$ has finite index.



Question: Is $H$ closed?



Problems:



I have a lot of trouble to understand these profinite groups in general. I know that they are inverse limits where every finite group is endowed with the discrete topology. Then we say that this inverse limit is endowed with the product topology of these discrete groups which I find hard already since I only worked with finite products before. I really have no clue how open/closed sets look like.



The only proof I understood is that in a compact topological space, every open subgroup is exactly the closed subgroups of finite index. I heard that there is some relation to the normal subgroups, but I did not really understand it. I need something to understand the proof in this post.



Could you please explain this concept to me? Thank you in advance!






abstract-algebra general-topology group-theory topological-groups normal-subgroups




abstract-algebra general-topology group-theory topological-groups normal-subgroups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 12:57







Diglett

















asked Jan 6 at 12:53









DiglettDiglett

9261521




9261521












  • $begingroup$
    Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
    $endgroup$
    – Yanko
    Jan 6 at 13:04




















  • $begingroup$
    Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
    $endgroup$
    – Yanko
    Jan 6 at 13:04


















$begingroup$
Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
$endgroup$
– Yanko
Jan 6 at 13:04






$begingroup$
Half an answer: If you assume that $G$ is finitely generated then every subgroup of finite index is open. math.wisc.edu/~boston/FLTJensen.pdf Theorem 1.1. I'll keep looking...
$endgroup$
– Yanko
Jan 6 at 13:04












1 Answer
1






active

oldest

votes


















0












$begingroup$

Please check out this paper page 4



https://arxiv.org/pdf/1406.6837.pdf



It says that the group $A_5^mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.



I can't find a concrete subgroup of $A_5^mathbb{N}$ which satisfies this but at least you know it's false.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
    $endgroup$
    – YCor
    Jan 11 at 2:21










  • $begingroup$
    @YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
    $endgroup$
    – Yanko
    Jan 11 at 12:03













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$begingroup$

Please check out this paper page 4



https://arxiv.org/pdf/1406.6837.pdf



It says that the group $A_5^mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.



I can't find a concrete subgroup of $A_5^mathbb{N}$ which satisfies this but at least you know it's false.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
    $endgroup$
    – YCor
    Jan 11 at 2:21










  • $begingroup$
    @YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
    $endgroup$
    – Yanko
    Jan 11 at 12:03


















0












$begingroup$

Please check out this paper page 4



https://arxiv.org/pdf/1406.6837.pdf



It says that the group $A_5^mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.



I can't find a concrete subgroup of $A_5^mathbb{N}$ which satisfies this but at least you know it's false.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
    $endgroup$
    – YCor
    Jan 11 at 2:21










  • $begingroup$
    @YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
    $endgroup$
    – Yanko
    Jan 11 at 12:03
















0












0








0





$begingroup$

Please check out this paper page 4



https://arxiv.org/pdf/1406.6837.pdf



It says that the group $A_5^mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.



I can't find a concrete subgroup of $A_5^mathbb{N}$ which satisfies this but at least you know it's false.






share|cite|improve this answer









$endgroup$



Please check out this paper page 4



https://arxiv.org/pdf/1406.6837.pdf



It says that the group $A_5^mathbb{N}$ is in "class 2", and they define class 2 to be compact groups having nonclosed subgroups of finite index.



I can't find a concrete subgroup of $A_5^mathbb{N}$ which satisfies this but at least you know it's false.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 13:08









YankoYanko

6,6841728




6,6841728








  • 1




    $begingroup$
    You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
    $endgroup$
    – YCor
    Jan 11 at 2:21










  • $begingroup$
    @YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
    $endgroup$
    – Yanko
    Jan 11 at 12:03
















  • 1




    $begingroup$
    You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
    $endgroup$
    – YCor
    Jan 11 at 2:21










  • $begingroup$
    @YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
    $endgroup$
    – Yanko
    Jan 11 at 12:03










1




1




$begingroup$
You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
$endgroup$
– YCor
Jan 11 at 2:21




$begingroup$
You won't find a concrete such subgroup: their existence highly relies on the choice axiom. Such subgroups (of $S^{mathbf{N}}$ with $S$ an arbitrary nontrivial finite group) can be found by picking a nonprincipal ultrafilter $omega$ on $mathbf{N}$, namely the normal subgroup ${(g_n)in S^{mathbf{N}}:lim_omega g_n=1}$.
$endgroup$
– YCor
Jan 11 at 2:21












$begingroup$
@YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
$endgroup$
– Yanko
Jan 11 at 12:03






$begingroup$
@YCor Thanks. Thinking about that it really makes sense now as measurable subgroups of compact groups of finite index are necessarily open (hence closed). So there could only be non-measurable counter-examples and usually the axiom of choice is the only way to construct non-measurable things.
$endgroup$
– Yanko
Jan 11 at 12:03




















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