$Spin_n(q)$, $SO_n(q)$, $Omega_n(q)$ and their projective images
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I am studying about the structure of orthogonal groups and struggling to understand the relations between groups in the title:
From algebraic groups point of view, it is known that $Spin_n(q)$ is the univesal cover of the simple orthogonal group $POmega_n(q)$, that is, $Spin_n(q)$ is the biggest perfect group such that $Spin_n(q)/Z(Spin_n(q)) cong POmega_n(q)$ (*).
Moreover, by the short exact sequence $1 rightarrow mathbb{Z}_2 rightarrow Spin_n(q) rightarrow SO_n(q) rightarrow 1 $, we have $Spin_n(q)/mathbb{Z}_2 cong SO_n(q)$ (**).
This is while $SO_n(q)$ is not generally a perfect group and it has a subgroup of index 2, which is its derived subgroup $Omega_n(q)$. Even more, the projective image of $SO_n(q)$ by its center is not generally a simple group. Therefore, using (**), we deduce that
$(Spin_n(q)/mathbb{Z}_2) / (Z(Spin_n(q))/mathbb{Z}_2) cong SO_n(q)/Z(SO_n(q))=PSO_n(q)$,
which is not generally simple. This seems to be in contradiction with (*).
Now my question is that how can we explain (*) and (**) together, for example in cases when $n$ is odd.
abstract-algebra finite-groups algebraic-groups groups-of-lie-type
asked Dec 5 '18 at 22:35
user97635user97635
423212
$endgroup$
add a comment |
$begingroup$
I am studying about the structure of orthogonal groups and struggling to understand the relations between groups in the title:
From algebraic groups point of view, it is known that $Spin_n(q)$ is the univesal cover of the simple orthogonal group $POmega_n(q)$, that is, $Spin_n(q)$ is the biggest perfect group such that $Spin_n(q)/Z(Spin_n(q)) cong POmega_n(q)$ (*).
Moreover, by the short exact sequence $1 rightarrow mathbb{Z}_2 rightarrow Spin_n(q) rightarrow SO_n(q) rightarrow 1 $, we have $Spin_n(q)/mathbb{Z}_2 cong SO_n(q)$ (**).
This is while $SO_n(q)$ is not generally a perfect group and it has a subgroup of index 2, which is its derived subgroup $Omega_n(q)$. Even more, the projective image of $SO_n(q)$ by its center is not generally a simple group. Therefore, using (**), we deduce that
$(Spin_n(q)/mathbb{Z}_2) / (Z(Spin_n(q))/mathbb{Z}_2) cong SO_n(q)/Z(SO_n(q))=PSO_n(q)$,
which is not generally simple. This seems to be in contradiction with (*).
Now my question is that how can we explain (*) and (**) together, for example in cases when $n$ is odd.
abstract-algebra finite-groups algebraic-groups groups-of-lie-type
asked Dec 5 '18 at 22:35
user97635user97635
423212
$endgroup$
$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
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@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
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@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
1
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16
add a comment |
$begingroup$
I am studying about the structure of orthogonal groups and struggling to understand the relations between groups in the title:
From algebraic groups point of view, it is known that $Spin_n(q)$ is the univesal cover of the simple orthogonal group $POmega_n(q)$, that is, $Spin_n(q)$ is the biggest perfect group such that $Spin_n(q)/Z(Spin_n(q)) cong POmega_n(q)$ (*).
Moreover, by the short exact sequence $1 rightarrow mathbb{Z}_2 rightarrow Spin_n(q) rightarrow SO_n(q) rightarrow 1 $, we have $Spin_n(q)/mathbb{Z}_2 cong SO_n(q)$ (**).
This is while $SO_n(q)$ is not generally a perfect group and it has a subgroup of index 2, which is its derived subgroup $Omega_n(q)$. Even more, the projective image of $SO_n(q)$ by its center is not generally a simple group. Therefore, using (**), we deduce that
$(Spin_n(q)/mathbb{Z}_2) / (Z(Spin_n(q))/mathbb{Z}_2) cong SO_n(q)/Z(SO_n(q))=PSO_n(q)$,
which is not generally simple. This seems to be in contradiction with (*).
Now my question is that how can we explain (*) and (**) together, for example in cases when $n$ is odd.
abstract-algebra finite-groups algebraic-groups groups-of-lie-type
asked Dec 5 '18 at 22:35
user97635user97635
423212
$endgroup$
I am studying about the structure of orthogonal groups and struggling to understand the relations between groups in the title:
From algebraic groups point of view, it is known that $Spin_n(q)$ is the univesal cover of the simple orthogonal group $POmega_n(q)$, that is, $Spin_n(q)$ is the biggest perfect group such that $Spin_n(q)/Z(Spin_n(q)) cong POmega_n(q)$ (*).
Moreover, by the short exact sequence $1 rightarrow mathbb{Z}_2 rightarrow Spin_n(q) rightarrow SO_n(q) rightarrow 1 $, we have $Spin_n(q)/mathbb{Z}_2 cong SO_n(q)$ (**).
This is while $SO_n(q)$ is not generally a perfect group and it has a subgroup of index 2, which is its derived subgroup $Omega_n(q)$. Even more, the projective image of $SO_n(q)$ by its center is not generally a simple group. Therefore, using (**), we deduce that
$(Spin_n(q)/mathbb{Z}_2) / (Z(Spin_n(q))/mathbb{Z}_2) cong SO_n(q)/Z(SO_n(q))=PSO_n(q)$,
which is not generally simple. This seems to be in contradiction with (*).
Now my question is that how can we explain (*) and (**) together, for example in cases when $n$ is odd.
abstract-algebra finite-groups algebraic-groups groups-of-lie-type
abstract-algebra finite-groups algebraic-groups groups-of-lie-type
asked Dec 5 '18 at 22:35
user97635user97635
423212
asked Dec 5 '18 at 22:35
user97635user97635
423212
asked Dec 5 '18 at 22:35
user97635user97635
423212
asked Dec 5 '18 at 22:35
user97635user97635
423212
asked Dec 5 '18 at 22:35
user97635user97635
423212
423212
$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
$begingroup$
@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
$begingroup$
@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
1
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16
add a comment |
$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
$begingroup$
@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
$begingroup$
@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
1
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16
$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
$begingroup$
@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
$begingroup$
@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
$begingroup$
@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
1
1
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16
add a comment |
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$begingroup$
How can $Spin(q)$ be perfect, and $SO(q)$ (a quotient) not be?
$endgroup$
– Hempelicious
Dec 5 '18 at 22:40
$begingroup$
@Hempelicious This is exactly what I am confusing about.
$endgroup$
– user97635
Dec 5 '18 at 22:57
$begingroup$
@Hempelicious Except a few cases, it is known that the universal covers of simple orthogonal groups are precisely the corresponding simply-connected algebraic group over the finite field. So here the universal cover must be Spin group which has SO as a quotient, see en.wikipedia.org/wiki/Spin_group
$endgroup$
– user97635
Dec 5 '18 at 23:05
$begingroup$
But $POmega_5(3)$ has order $25920$, while $SO_5(3)$ has order twice that. Since the universal cover of $POmega_5(3)$ is 2-sheeted, how can both coverings be the same group? The orders don't match.
$endgroup$
– Hempelicious
Dec 6 '18 at 5:38
1
$begingroup$
According to this, the map you have in (**) is not surjective when $q$ is odd.
$endgroup$
– Hempelicious
Dec 6 '18 at 6:16