Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$?
$begingroup$
Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?
In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?
I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?
Thank you so much in advance.
group-theory representation-theory
asked Jan 2 at 16:11
mathstudentmathstudent
394
$endgroup$
add a comment |
$begingroup$
Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?
In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?
I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?
Thank you so much in advance.
group-theory representation-theory
asked Jan 2 at 16:11
mathstudentmathstudent
394
$endgroup$
add a comment |
$begingroup$
Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?
In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?
I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?
Thank you so much in advance.
group-theory representation-theory
asked Jan 2 at 16:11
mathstudentmathstudent
394
$endgroup$
Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?
In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?
I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?
Thank you so much in advance.
group-theory representation-theory
group-theory representation-theory
asked Jan 2 at 16:11
mathstudentmathstudent
394
asked Jan 2 at 16:11
mathstudentmathstudent
394
edited Jan 3 at 8:06
mathstudent
asked Jan 2 at 16:11
mathstudentmathstudent
394
asked Jan 2 at 16:11
mathstudentmathstudent
394
asked Jan 2 at 16:11
mathstudentmathstudent
394
394
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
(this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
Bbb Z^{n-1})
=GL(n-1,Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
|
show 4 more comments
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$begingroup$
For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
(this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
Bbb Z^{n-1})
=GL(n-1,Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
|
show 4 more comments
$begingroup$
For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
(this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
Bbb Z^{n-1})
=GL(n-1,Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
|
show 4 more comments
$begingroup$
For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
(this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
Bbb Z^{n-1})
=GL(n-1,Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.
Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
(this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
Bbb Z^{n-1})
=GL(n-1,Bbb Z)$.
In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 2 at 16:17
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
|
show 4 more comments
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
$endgroup$
– mathstudent
Jan 2 at 17:00
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
What are your favourite generators of $A_5$? @mathstudent
$endgroup$
– Lord Shark the Unknown
Jan 2 at 17:02
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
$begingroup$
Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
$endgroup$
– mathstudent
Jan 2 at 17:04
1
1
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
$begingroup$
It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
$endgroup$
– Derek Holt
Jan 3 at 8:28
1
1
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
$begingroup$
I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
$endgroup$
– Derek Holt
Jan 3 at 9:08
|
show 4 more comments
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