Mixing
A discrete normal subgroup is contained in the center
$begingroup$
Let $G$ be a connected Lie group and $N$ a discrete normal subgroup of $G$. Then $N$ is contained in the center $Z(G)$.
I've fooled around with this for a little bit and I can't figure out how to use the hypothesis that $N$ is discrete. I think maybe it uses some fact that I do not know.
abstract-algebra group-theory lie-groups
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
$endgroup$
add a comment |
$begingroup$
Let $G$ be a connected Lie group and $N$ a discrete normal subgroup of $G$. Then $N$ is contained in the center $Z(G)$.
I've fooled around with this for a little bit and I can't figure out how to use the hypothesis that $N$ is discrete. I think maybe it uses some fact that I do not know.
abstract-algebra group-theory lie-groups
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
$endgroup$
$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46
add a comment |
$begingroup$
Let $G$ be a connected Lie group and $N$ a discrete normal subgroup of $G$. Then $N$ is contained in the center $Z(G)$.
I've fooled around with this for a little bit and I can't figure out how to use the hypothesis that $N$ is discrete. I think maybe it uses some fact that I do not know.
abstract-algebra group-theory lie-groups
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
$endgroup$
Let $G$ be a connected Lie group and $N$ a discrete normal subgroup of $G$. Then $N$ is contained in the center $Z(G)$.
I've fooled around with this for a little bit and I can't figure out how to use the hypothesis that $N$ is discrete. I think maybe it uses some fact that I do not know.
abstract-algebra group-theory lie-groups
abstract-algebra group-theory lie-groups
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
edited Jan 27 at 19:45
TuoTuo
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
asked Jan 27 at 18:03
TuoTuoTuoTuo
1,781516
1,781516
$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46
add a comment |
$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46
$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46
$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $n$ be an element of the subgroup. Then the function $f:Gto N$ sending $gmapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $n$ be an element of the subgroup. Then the function $f:Gto N$ sending $gmapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
add a comment |
$begingroup$
Let $n$ be an element of the subgroup. Then the function $f:Gto N$ sending $gmapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
$endgroup$
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
add a comment |
$begingroup$
Let $n$ be an element of the subgroup. Then the function $f:Gto N$ sending $gmapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
$endgroup$
Let $n$ be an element of the subgroup. Then the function $f:Gto N$ sending $gmapsto gng^{-1}$ is continuous, and since $G$ is connected, so is the image of $f$. Since $N$ is discrete, $f(g) =n$ for all $g$.
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
edited Jan 27 at 19:41
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
answered Jan 27 at 18:12
Matt SamuelMatt Samuel
39k63769
39k63769
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
add a comment |
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
Ah yes, I forgot to say $G$ was connected. That's one of my permanent assumptions.
$endgroup$
– TuoTuo
Jan 27 at 19:40
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
$begingroup$
@Tuo I'll edit my answer then.
$endgroup$
– Matt Samuel
Jan 27 at 19:41
add a comment |
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$begingroup$
Re: your initially-missing "connectedness" hypothesis: It's worth remembering that every finite group is a Lie Group (0-dimensional, discrete topology), so when you come to prove a fact like this, and you notice it's not true for finite groups, then surely connectedness will come into play!
$endgroup$
– John Hughes
Jan 27 at 19:46