Does the Lie group $G_2$ contain any normal subgroups?
$begingroup$
A nonabelian Lie group is called a simple Lie group if it contains no nontrivial connected normal subgroups. On the other hand, a group is called simple if it contains no nontrivial normal subgroups. So a simple Lie group may not be simple as a group. For example, $SU(2)$ is a simple Lie group, but it is not a simple group because ${pm I}$ is a normal (actually, central) subgroup.
The $14$-dimensional compact Lie group $G_2$ is a simple Lie group. Is $G_2$ a simple group? That is:
- Does $G_2$ contain a nontrivial normal subgroup (which is necessarily disconnected)?
I am aware that $G_2$ has trivial center, so any such normal subgroup is not central.
Actually, my real question (which may be easier to answer) is:
- Does $G_2$ contain a normal subgroup isomorphic to $mathbb{Z}_2$?
As $G_2$ has rank two, it has a subgroup isomorphic to $S^1times S^1$ and therefore does have subgroups isomorphic to $mathbb{Z}_2$, but I don't know if any of them are normal.
lie-groups normal-subgroups
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
add a comment |
$begingroup$
A nonabelian Lie group is called a simple Lie group if it contains no nontrivial connected normal subgroups. On the other hand, a group is called simple if it contains no nontrivial normal subgroups. So a simple Lie group may not be simple as a group. For example, $SU(2)$ is a simple Lie group, but it is not a simple group because ${pm I}$ is a normal (actually, central) subgroup.
The $14$-dimensional compact Lie group $G_2$ is a simple Lie group. Is $G_2$ a simple group? That is:
- Does $G_2$ contain a nontrivial normal subgroup (which is necessarily disconnected)?
I am aware that $G_2$ has trivial center, so any such normal subgroup is not central.
Actually, my real question (which may be easier to answer) is:
- Does $G_2$ contain a normal subgroup isomorphic to $mathbb{Z}_2$?
As $G_2$ has rank two, it has a subgroup isomorphic to $S^1times S^1$ and therefore does have subgroups isomorphic to $mathbb{Z}_2$, but I don't know if any of them are normal.
lie-groups normal-subgroups
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
add a comment |
$begingroup$
A nonabelian Lie group is called a simple Lie group if it contains no nontrivial connected normal subgroups. On the other hand, a group is called simple if it contains no nontrivial normal subgroups. So a simple Lie group may not be simple as a group. For example, $SU(2)$ is a simple Lie group, but it is not a simple group because ${pm I}$ is a normal (actually, central) subgroup.
The $14$-dimensional compact Lie group $G_2$ is a simple Lie group. Is $G_2$ a simple group? That is:
- Does $G_2$ contain a nontrivial normal subgroup (which is necessarily disconnected)?
I am aware that $G_2$ has trivial center, so any such normal subgroup is not central.
Actually, my real question (which may be easier to answer) is:
- Does $G_2$ contain a normal subgroup isomorphic to $mathbb{Z}_2$?
As $G_2$ has rank two, it has a subgroup isomorphic to $S^1times S^1$ and therefore does have subgroups isomorphic to $mathbb{Z}_2$, but I don't know if any of them are normal.
lie-groups normal-subgroups
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
$endgroup$
A nonabelian Lie group is called a simple Lie group if it contains no nontrivial connected normal subgroups. On the other hand, a group is called simple if it contains no nontrivial normal subgroups. So a simple Lie group may not be simple as a group. For example, $SU(2)$ is a simple Lie group, but it is not a simple group because ${pm I}$ is a normal (actually, central) subgroup.
The $14$-dimensional compact Lie group $G_2$ is a simple Lie group. Is $G_2$ a simple group? That is:
- Does $G_2$ contain a nontrivial normal subgroup (which is necessarily disconnected)?
I am aware that $G_2$ has trivial center, so any such normal subgroup is not central.
Actually, my real question (which may be easier to answer) is:
- Does $G_2$ contain a normal subgroup isomorphic to $mathbb{Z}_2$?
As $G_2$ has rank two, it has a subgroup isomorphic to $S^1times S^1$ and therefore does have subgroups isomorphic to $mathbb{Z}_2$, but I don't know if any of them are normal.
lie-groups normal-subgroups
lie-groups normal-subgroups
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
asked Dec 1 '18 at 18:13
Michael AlbaneseMichael Albanese
63.2k1598303
63.2k1598303
add a comment |
add a comment |
1 Answer
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$begingroup$
If there is a disconnected normal subgroup $N$ of $G_2$, then its connected component containing the identity (call it $N_0$) is a connected normal subgroup. Since $G_2$ is simple (as a Lie group), it follows that $N_0$ is the identity. For every $n in N$, $n N_0 = { n }$ is a connected component of $N$. Hence $N$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $n$, the map $G to N$ sending $g mapsto gng^{-1}$ is a connected subset of $N$ containing $n$, and therefore consists of only the point $n$).
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
$endgroup$
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
add a comment |
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$begingroup$
If there is a disconnected normal subgroup $N$ of $G_2$, then its connected component containing the identity (call it $N_0$) is a connected normal subgroup. Since $G_2$ is simple (as a Lie group), it follows that $N_0$ is the identity. For every $n in N$, $n N_0 = { n }$ is a connected component of $N$. Hence $N$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $n$, the map $G to N$ sending $g mapsto gng^{-1}$ is a connected subset of $N$ containing $n$, and therefore consists of only the point $n$).
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
$endgroup$
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
add a comment |
$begingroup$
If there is a disconnected normal subgroup $N$ of $G_2$, then its connected component containing the identity (call it $N_0$) is a connected normal subgroup. Since $G_2$ is simple (as a Lie group), it follows that $N_0$ is the identity. For every $n in N$, $n N_0 = { n }$ is a connected component of $N$. Hence $N$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $n$, the map $G to N$ sending $g mapsto gng^{-1}$ is a connected subset of $N$ containing $n$, and therefore consists of only the point $n$).
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
$endgroup$
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
add a comment |
$begingroup$
If there is a disconnected normal subgroup $N$ of $G_2$, then its connected component containing the identity (call it $N_0$) is a connected normal subgroup. Since $G_2$ is simple (as a Lie group), it follows that $N_0$ is the identity. For every $n in N$, $n N_0 = { n }$ is a connected component of $N$. Hence $N$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $n$, the map $G to N$ sending $g mapsto gng^{-1}$ is a connected subset of $N$ containing $n$, and therefore consists of only the point $n$).
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
$endgroup$
If there is a disconnected normal subgroup $N$ of $G_2$, then its connected component containing the identity (call it $N_0$) is a connected normal subgroup. Since $G_2$ is simple (as a Lie group), it follows that $N_0$ is the identity. For every $n in N$, $n N_0 = { n }$ is a connected component of $N$. Hence $N$ is discrete, and therefore central (a discrete normal subgroup of a connected Lie group is central: for a fixed $n$, the map $G to N$ sending $g mapsto gng^{-1}$ is a connected subset of $N$ containing $n$, and therefore consists of only the point $n$).
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
edited Dec 1 '18 at 19:41
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
answered Dec 1 '18 at 19:27
TedTed
21.5k13260
21.5k13260
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
add a comment |
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
$begingroup$
All I was missing is the fact that a discrete normal subgroup of a connected Lie group is central. Thanks!
$endgroup$
– Michael Albanese
Dec 1 '18 at 21:03
add a comment |
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