Discrete quotient group from closed subgroup

[etnad179]

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!

 

[micromass]

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.

 

[etnad179]

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.

Hi thanks for your quick reply. I'm still a little confused - the singletons of G/H are open is fine - but why does that imply the topology on G/H is discrete?

 

[micromass]

Because every set is the union of singletons. And the union of open sets is open...

 

[etnad179]

Because every set is the union of singletons. And the union of open sets is open...

Sorry, I'm quite new to this - I might just be missing a definition... but why is the topological group discrete if the set is open?

 

[micromass]

OK, how did you define discrete?

 

[etnad179]

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.

 

[micromass]

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.

Ah, that is probably where the confusion is. A topological space is called discrete if all sets are open. Of course, a discrete space is a totally disconnected space, but not conversely.

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.

 

[etnad179]

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!

 

[micromass]

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.

 

[etnad179]

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