Mixing
If $G subset GL_n(mathbb{C})$ has finitely many conjugacy classes for all element, is it finite?
$begingroup$
Let be $G subset GL_n(mathbb{C})$ such that there is some $r in mathbb{N}^{*}$ and $g_1, ldots, g_r in G$ so that for all $g in G$, $g$ is conjugated to some $g_i, i in [[1, r]]$ in $G$.
Is $G$ finite? I feel like that yes, I tried to poke the stabilizer, but I have no reason to think that it'd be finite.
group-theory representation-theory
asked Jan 27 at 11:16
RaitoRaito
686415
$endgroup$
|
show 1 more comment
$begingroup$
Let be $G subset GL_n(mathbb{C})$ such that there is some $r in mathbb{N}^{*}$ and $g_1, ldots, g_r in G$ so that for all $g in G$, $g$ is conjugated to some $g_i, i in [[1, r]]$ in $G$.
Is $G$ finite? I feel like that yes, I tried to poke the stabilizer, but I have no reason to think that it'd be finite.
group-theory representation-theory
asked Jan 27 at 11:16
RaitoRaito
686415
$endgroup$
3
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
2
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
2
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
1
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
2
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26
|
show 1 more comment
$begingroup$
Let be $G subset GL_n(mathbb{C})$ such that there is some $r in mathbb{N}^{*}$ and $g_1, ldots, g_r in G$ so that for all $g in G$, $g$ is conjugated to some $g_i, i in [[1, r]]$ in $G$.
Is $G$ finite? I feel like that yes, I tried to poke the stabilizer, but I have no reason to think that it'd be finite.
group-theory representation-theory
asked Jan 27 at 11:16
RaitoRaito
686415
$endgroup$
Let be $G subset GL_n(mathbb{C})$ such that there is some $r in mathbb{N}^{*}$ and $g_1, ldots, g_r in G$ so that for all $g in G$, $g$ is conjugated to some $g_i, i in [[1, r]]$ in $G$.
Is $G$ finite? I feel like that yes, I tried to poke the stabilizer, but I have no reason to think that it'd be finite.
group-theory representation-theory
group-theory representation-theory
asked Jan 27 at 11:16
RaitoRaito
686415
asked Jan 27 at 11:16
RaitoRaito
686415
edited Jan 27 at 12:00
Raito
asked Jan 27 at 11:16
RaitoRaito
686415
asked Jan 27 at 11:16
RaitoRaito
686415
asked Jan 27 at 11:16
RaitoRaito
686415
686415
3
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
2
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
2
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
1
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
2
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26
|
show 1 more comment
3
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
2
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
2
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
1
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
2
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26
3
3
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
2
2
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
2
2
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
1
1
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
2
2
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).
Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.
The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''in G$.
If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''in G$, i.e., $G={1}$.
In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.
In other words, this proves that if $G$ is a subgroup of $mathrm{GL}(n,mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.
If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $Hneq{1}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.
edited Jan 31 at 10:13
Max
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
$endgroup$
add a comment |
$begingroup$
Yes, $G$ must be finite.
There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).
But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
$endgroup$
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
add a comment |
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2 Answers
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2 Answers
2
active
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active
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$begingroup$
$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).
Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.
The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''in G$.
If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''in G$, i.e., $G={1}$.
In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.
In other words, this proves that if $G$ is a subgroup of $mathrm{GL}(n,mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.
If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $Hneq{1}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.
edited Jan 31 at 10:13
Max
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
$endgroup$
add a comment |
$begingroup$
$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).
Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.
The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''in G$.
If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''in G$, i.e., $G={1}$.
In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.
In other words, this proves that if $G$ is a subgroup of $mathrm{GL}(n,mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.
If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $Hneq{1}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.
edited Jan 31 at 10:13
Max
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
$endgroup$
add a comment |
$begingroup$
$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).
Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.
The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''in G$.
If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''in G$, i.e., $G={1}$.
In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.
In other words, this proves that if $G$ is a subgroup of $mathrm{GL}(n,mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.
If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $Hneq{1}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.
edited Jan 31 at 10:13
Max
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
$endgroup$
$G$ must be finite. Here's a proof working for $G$ arbitrary (Derek's proof works only when $G$ is finitely generated).
Passing to a finite index subgroup, we can suppose that $G$ has connected Zariski closure.
The assumption implies that $G$ has finitely many traces, and using connectedness of the Zariski closure, it has a single trace. So $mathrm{Tr}(g(g'-g''))=0$ for all $g,g',g''in G$.
If we assume in addition that $G$ is irreducible and hence linearly generates the space of matrices, since the Trace yields a nondegenerate bilinear form, we deduce that $g'-g''=0$ for all $g',g''in G$, i.e., $G={1}$.
In general, this applies to the projection on all irreducible diagonal blocks, which are thus one-dimensional.
In other words, this proves that if $G$ is a subgroup of $mathrm{GL}(n,mathbf{C})$ with finitely many traces, then $G$ has a finite index subgroup $H$ (namely, the intersection with the identity component of its Zariski closure) that is conjugated to a group of upper triangular unipotent matrices.
If $G$ has finitely many conjugacy classes, so does $H$. Then $H$ has a normal series in which all subquotients are torsion-free abelian groups (it inherits this from the group of unipotent upper triangular matrices). This implies that if $Hneq{1}$ then it has an infinite abelian quotient (namely a nontrivial torsion-free abelian quotient). This implies that $H$ has an infinite number of conjugacy classes.
edited Jan 31 at 10:13
Max
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
edited Jan 31 at 10:13
Max
15.6k11143
edited Jan 31 at 10:13
Max
15.6k11143
edited Jan 31 at 10:13
Max
15.6k11143
15.6k11143
answered Jan 31 at 6:20
YCorYCor
8,2671129
answered Jan 31 at 6:20
YCorYCor
8,2671129
answered Jan 31 at 6:20
YCorYCor
8,2671129
8,2671129
add a comment |
add a comment |
$begingroup$
Yes, $G$ must be finite.
There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).
But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
$endgroup$
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
add a comment |
$begingroup$
Yes, $G$ must be finite.
There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).
But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
$endgroup$
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
add a comment |
$begingroup$
Yes, $G$ must be finite.
There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).
But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
$endgroup$
Yes, $G$ must be finite.
There are only finitely many finite groups (up to isomorphism of course) with a bounded number of conjugacy classes (see here for a proof).
But it is well known that linear groups are residually finite, and so if there were an infinite example, then it would have arbitrarily large finite quotients, each with only $r$ conjugacy classes.
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
answered Jan 27 at 12:07
Derek HoltDerek Holt
54.3k53573
54.3k53573
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
add a comment |
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
$begingroup$
"Linear groups are residually finite": this is only true for finitely generated groups.
$endgroup$
– YCor
Jan 31 at 6:07
add a comment |
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3
$begingroup$
Is the conjugation by elements of $G$ or by elements of $mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:21
2
$begingroup$
If conjugacy by elements of $GL_n$, consider the group of all $begin{pmatrix}1&z\0&1end{pmatrix}$, which is isomoirphic to $Bbb C$.
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:23
2
$begingroup$
@Raito that's still a bit unclear. Certainly the $g_i$ are in $G$. By assumption, if $gin G$, then $g = hg_i h^{-1}$ for some $h$. My question is: is $hin G$, or is $hin mathrm{GL}_n$?
$endgroup$
– Mathmo123
Jan 27 at 11:25
1
$begingroup$
@Raito you said that $g$ and the $g_i$ are in $G$, you also said they are conjugated, i.e., that $g=hg_ih^{-1}$ for some $h$, but Mathmo's question is whether $hin G$
$endgroup$
– Hagen von Eitzen
Jan 27 at 11:25
2
$begingroup$
@HagenvonEitzen Right, sorry, the conjugacy is in G.
$endgroup$
– Raito
Jan 27 at 11:26