
#1
Mar2811, 12:01 AM

P: 11

Hi All,
I've come across a theorem that I'm trying to prove, which states that: The quotient group G/H is a discrete group iff the normal subgroup H is open. In fact I'm only really interested in the direction H open implies G/H discrete.. To a lesser extent I'm also interested in the H being closed iff G/H Haussdorf. Thanks! 



#2
Mar2811, 08:57 AM

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P: 16,703

for every g holds that gH is an open set. From the definition of the quotient topology, we get that {gH} is an open set. Thus the singletons of G/H are open, this means that the topology on G/H is discrete.




#3
Mar2811, 01:49 PM

P: 11





#4
Mar2811, 01:52 PM

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P: 16,703

Discrete quotient group from closed subgroup
Because every set is the union of singletons. And the union of open sets is open...




#5
Mar2811, 01:59 PM

P: 11





#7
Mar2811, 02:14 PM

P: 11

I took the definition of discrete group to be one that is totally disconnected...
I would have thought that if the set of singletons was open, one could always go to another group element by following some connected path? If the singletons were closed, I would have guessed that then you would have a discrete group... I'm obviously thinking about this completely incorrectly though. 



#8
Mar2811, 02:21 PM

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P: 16,703

So, to show that a G/H is a discrete space, it suffices to show that all sets are open. This is what I just did. 



#9
Mar2811, 02:53 PM

P: 11

Hmmm  that makes some sense.
So my definition of a discrete group (being one that is totally disconnect) is incomplete? Also, would you know how this idea of open sets links with what physicists call discrete groups? i.e. groups with discrete elements? Thanks so much! 



#10
Mar2811, 02:58 PM

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P: 16,703

I can't comment on what physicists mean with discrete, since I know virtually nothing about physics
But all I know that a topological space is called discrete if all sets are open. So I guess that it would be natural to call a topological group discrete if all sets are open... So yes, I fear that you're working with the wrong definitions here. 



#11
Mar2811, 02:59 PM

P: 11

Oh well, thanks for the help! Really appreciate it!



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