What finite non-abelian group is generated by $operatorname{diag}(1,w,w^2,ldots,w^{N-1})$ (with $w=e^{2pi...
$begingroup$
What is the finite nonabelian group? generated by the following elements and satifies the rules:
$$A=left(begin{array}{ccccc}
1&0&0&cdots&0\
0&omega&0&cdots&0\
0&0&omega^2&cdots &0\
vdots&vdots&vdots& &vdots\
0&0&0&cdots&omega^{N-1}
end{array}right)$$ where $omega^{N}=1$ and $omega = e^{2pi i/N}$.
$$B=left(begin{array}{ccccc}
0&1&0&cdots&0\
0&0&1&cdots&0\
vdots&vdots &vdots & ddots &vdots\
0&0&0&cdots&1\
1&0&0&cdots&0
end{array}right)$$
$$AB=omega ;BA$$ (or $AB=omega^{-1} ;BA$)
It looks like this nonabelian group has an order $N^3$ at least. Because
$$
A^N=B^N=omega^N=1,
$$
and all elements of these are distinct.
When $N=2$, the answer of this nonabelian group seems to be a quaternion group of order $2^3=8$.
abstract-algebra group-theory finite-groups finitely-generated
edited Dec 21 '18 at 17:07
Shaun
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
$endgroup$
|
show 1 more comment
$begingroup$
What is the finite nonabelian group? generated by the following elements and satifies the rules:
$$A=left(begin{array}{ccccc}
1&0&0&cdots&0\
0&omega&0&cdots&0\
0&0&omega^2&cdots &0\
vdots&vdots&vdots& &vdots\
0&0&0&cdots&omega^{N-1}
end{array}right)$$ where $omega^{N}=1$ and $omega = e^{2pi i/N}$.
$$B=left(begin{array}{ccccc}
0&1&0&cdots&0\
0&0&1&cdots&0\
vdots&vdots &vdots & ddots &vdots\
0&0&0&cdots&1\
1&0&0&cdots&0
end{array}right)$$
$$AB=omega ;BA$$ (or $AB=omega^{-1} ;BA$)
It looks like this nonabelian group has an order $N^3$ at least. Because
$$
A^N=B^N=omega^N=1,
$$
and all elements of these are distinct.
When $N=2$, the answer of this nonabelian group seems to be a quaternion group of order $2^3=8$.
abstract-algebra group-theory finite-groups finitely-generated
edited Dec 21 '18 at 17:07
Shaun
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
$endgroup$
1
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
1
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
1
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55
|
show 1 more comment
$begingroup$
What is the finite nonabelian group? generated by the following elements and satifies the rules:
$$A=left(begin{array}{ccccc}
1&0&0&cdots&0\
0&omega&0&cdots&0\
0&0&omega^2&cdots &0\
vdots&vdots&vdots& &vdots\
0&0&0&cdots&omega^{N-1}
end{array}right)$$ where $omega^{N}=1$ and $omega = e^{2pi i/N}$.
$$B=left(begin{array}{ccccc}
0&1&0&cdots&0\
0&0&1&cdots&0\
vdots&vdots &vdots & ddots &vdots\
0&0&0&cdots&1\
1&0&0&cdots&0
end{array}right)$$
$$AB=omega ;BA$$ (or $AB=omega^{-1} ;BA$)
It looks like this nonabelian group has an order $N^3$ at least. Because
$$
A^N=B^N=omega^N=1,
$$
and all elements of these are distinct.
When $N=2$, the answer of this nonabelian group seems to be a quaternion group of order $2^3=8$.
abstract-algebra group-theory finite-groups finitely-generated
edited Dec 21 '18 at 17:07
Shaun
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
$endgroup$
What is the finite nonabelian group? generated by the following elements and satifies the rules:
$$A=left(begin{array}{ccccc}
1&0&0&cdots&0\
0&omega&0&cdots&0\
0&0&omega^2&cdots &0\
vdots&vdots&vdots& &vdots\
0&0&0&cdots&omega^{N-1}
end{array}right)$$ where $omega^{N}=1$ and $omega = e^{2pi i/N}$.
$$B=left(begin{array}{ccccc}
0&1&0&cdots&0\
0&0&1&cdots&0\
vdots&vdots &vdots & ddots &vdots\
0&0&0&cdots&1\
1&0&0&cdots&0
end{array}right)$$
$$AB=omega ;BA$$ (or $AB=omega^{-1} ;BA$)
It looks like this nonabelian group has an order $N^3$ at least. Because
$$
A^N=B^N=omega^N=1,
$$
and all elements of these are distinct.
When $N=2$, the answer of this nonabelian group seems to be a quaternion group of order $2^3=8$.
abstract-algebra group-theory finite-groups finitely-generated
abstract-algebra group-theory finite-groups finitely-generated
edited Dec 21 '18 at 17:07
Shaun
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
edited Dec 21 '18 at 17:07
Shaun
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
edited Dec 21 '18 at 17:07
Shaun
9,870113684
edited Dec 21 '18 at 17:07
Shaun
9,870113684
edited Dec 21 '18 at 17:07
Shaun
9,870113684
9,870113684
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
asked Dec 19 '18 at 2:49
annie heartannie heart
668721
668721
1
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
1
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
1
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55
|
show 1 more comment
1
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
1
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
1
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55
1
1
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
1
1
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
1
1
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The group $G = langle A,B rangle$ generated by $A$ and $B$ has order $N^3$ for all $N>1$.
There are $N$ conjugates $A_i = A^{B^i}$ $(0 le i < N)$ of $A$ under powers of $B$, but $A_iA_{i-1}^{-1} = omega I_n$, so the $A_i$ generate the group $N = langle A,omega I_n rangle$, which has order $N^2$. Then $N lhd G$ and $G/N$ is generated by the image $BN$ of $B$ and is cyclic of order $N$.
This is a nilpotent group of class $2$ with $Z(G) = [G,G] = langle omega I_n rangle$.
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
$endgroup$
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The group $G = langle A,B rangle$ generated by $A$ and $B$ has order $N^3$ for all $N>1$.
There are $N$ conjugates $A_i = A^{B^i}$ $(0 le i < N)$ of $A$ under powers of $B$, but $A_iA_{i-1}^{-1} = omega I_n$, so the $A_i$ generate the group $N = langle A,omega I_n rangle$, which has order $N^2$. Then $N lhd G$ and $G/N$ is generated by the image $BN$ of $B$ and is cyclic of order $N$.
This is a nilpotent group of class $2$ with $Z(G) = [G,G] = langle omega I_n rangle$.
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
$endgroup$
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
add a comment |
$begingroup$
The group $G = langle A,B rangle$ generated by $A$ and $B$ has order $N^3$ for all $N>1$.
There are $N$ conjugates $A_i = A^{B^i}$ $(0 le i < N)$ of $A$ under powers of $B$, but $A_iA_{i-1}^{-1} = omega I_n$, so the $A_i$ generate the group $N = langle A,omega I_n rangle$, which has order $N^2$. Then $N lhd G$ and $G/N$ is generated by the image $BN$ of $B$ and is cyclic of order $N$.
This is a nilpotent group of class $2$ with $Z(G) = [G,G] = langle omega I_n rangle$.
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
$endgroup$
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
add a comment |
$begingroup$
The group $G = langle A,B rangle$ generated by $A$ and $B$ has order $N^3$ for all $N>1$.
There are $N$ conjugates $A_i = A^{B^i}$ $(0 le i < N)$ of $A$ under powers of $B$, but $A_iA_{i-1}^{-1} = omega I_n$, so the $A_i$ generate the group $N = langle A,omega I_n rangle$, which has order $N^2$. Then $N lhd G$ and $G/N$ is generated by the image $BN$ of $B$ and is cyclic of order $N$.
This is a nilpotent group of class $2$ with $Z(G) = [G,G] = langle omega I_n rangle$.
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
$endgroup$
The group $G = langle A,B rangle$ generated by $A$ and $B$ has order $N^3$ for all $N>1$.
There are $N$ conjugates $A_i = A^{B^i}$ $(0 le i < N)$ of $A$ under powers of $B$, but $A_iA_{i-1}^{-1} = omega I_n$, so the $A_i$ generate the group $N = langle A,omega I_n rangle$, which has order $N^2$. Then $N lhd G$ and $G/N$ is generated by the image $BN$ of $B$ and is cyclic of order $N$.
This is a nilpotent group of class $2$ with $Z(G) = [G,G] = langle omega I_n rangle$.
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
answered Dec 19 '18 at 8:32
Derek HoltDerek Holt
54.4k53574
54.4k53574
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
add a comment |
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
$begingroup$
Thanks! +1, Is there a name of this group?
$endgroup$
– annie heart
Dec 19 '18 at 15:39
2
2
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
$begingroup$
@annie Heisenberg group modulo N no good?
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 20:33
1
1
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
$begingroup$
@CosmasZachos Yes, that's right! Or the Heisenberg group over the ring ${mathbb Z}/N{mathbb Z}$.
$endgroup$
– Derek Holt
Dec 19 '18 at 22:34
add a comment |
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1
$begingroup$
If you're defining $omega$ to be a root of unity, isn't condition (3) redundant? Also, (2) isn't really a condition; it's just the definition of $B$. Also, where does this question come from? What's the context of this exercise?
$endgroup$
– Mike Pierce
Dec 19 '18 at 3:44
$begingroup$
It is just something I read from a paper. It is not homework.
$endgroup$
– annie heart
Dec 19 '18 at 3:48
$begingroup$
They did not discuss the group structure - I wonder what it is? Shall be $mathbb{Z}_N^2$-extension over $mathbb{Z}_N$; or the other way around
$endgroup$
– annie heart
Dec 19 '18 at 3:49
1
$begingroup$
Compare to this.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 4:07
1
$begingroup$
For N=3, the Nonion group was discovered in 1884 by J J Sylvester, along with the generalization to the clock and shift matrices you are looking at.
$endgroup$
– Cosmas Zachos
Dec 19 '18 at 13:55