What finite non-abelian group is generated by $operatorname{diag}(1,w,w^2,ldots,w^{N-1})$ (with $w=e^{2pi...












7












$begingroup$



What is the finite nonabelian group? generated by the following elements and satifies the rules:




  1. $$A=left(begin{array}{ccccc}
    1&0&0&cdots&0\
    0&omega&0&cdots&0\
    0&0&omega^2&cdots &0\
    vdots&vdots&vdots& &vdots\
    0&0&0&cdots&omega^{N-1}
    end{array}right)$$     where $omega^{N}=1$ and $omega = e^{2pi i/N}$.


  2. $$B=left(begin{array}{ccccc}
    0&1&0&cdots&0\
    0&0&1&cdots&0\
    vdots&vdots &vdots & ddots &vdots\
    0&0&0&cdots&1\
    1&0&0&cdots&0
    end{array}right)$$


  3. $$AB=omega ;BA$$ (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







share|cite|improve this question











$endgroup$








  • 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
















7












$begingroup$



What is the finite nonabelian group? generated by the following elements and satifies the rules:




  1. $$A=left(begin{array}{ccccc}
    1&0&0&cdots&0\
    0&omega&0&cdots&0\
    0&0&omega^2&cdots &0\
    vdots&vdots&vdots& &vdots\
    0&0&0&cdots&omega^{N-1}
    end{array}right)$$     where $omega^{N}=1$ and $omega = e^{2pi i/N}$.


  2. $$B=left(begin{array}{ccccc}
    0&1&0&cdots&0\
    0&0&1&cdots&0\
    vdots&vdots &vdots & ddots &vdots\
    0&0&0&cdots&1\
    1&0&0&cdots&0
    end{array}right)$$


  3. $$AB=omega ;BA$$ (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







share|cite|improve this question











$endgroup$








  • 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














7












7








7


3



$begingroup$



What is the finite nonabelian group? generated by the following elements and satifies the rules:




  1. $$A=left(begin{array}{ccccc}
    1&0&0&cdots&0\
    0&omega&0&cdots&0\
    0&0&omega^2&cdots &0\
    vdots&vdots&vdots& &vdots\
    0&0&0&cdots&omega^{N-1}
    end{array}right)$$     where $omega^{N}=1$ and $omega = e^{2pi i/N}$.


  2. $$B=left(begin{array}{ccccc}
    0&1&0&cdots&0\
    0&0&1&cdots&0\
    vdots&vdots &vdots & ddots &vdots\
    0&0&0&cdots&1\
    1&0&0&cdots&0
    end{array}right)$$


  3. $$AB=omega ;BA$$ (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







share|cite|improve this question











$endgroup$





What is the finite nonabelian group? generated by the following elements and satifies the rules:




  1. $$A=left(begin{array}{ccccc}
    1&0&0&cdots&0\
    0&omega&0&cdots&0\
    0&0&omega^2&cdots &0\
    vdots&vdots&vdots& &vdots\
    0&0&0&cdots&omega^{N-1}
    end{array}right)$$     where $omega^{N}=1$ and $omega = e^{2pi i/N}$.


  2. $$B=left(begin{array}{ccccc}
    0&1&0&cdots&0\
    0&0&1&cdots&0\
    vdots&vdots &vdots & ddots &vdots\
    0&0&0&cdots&1\
    1&0&0&cdots&0
    end{array}right)$$


  3. $$AB=omega ;BA$$ (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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 17:07









Shaun

9,870113684




9,870113684










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














  • 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










1 Answer
1






active

oldest

votes


















3












$begingroup$

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 $Z(G) = [G,G] = langle omega I_n rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

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 $Z(G) = [G,G] = langle omega I_n rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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 $Z(G) = [G,G] = langle omega I_n rangle$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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 $Z(G) = [G,G] = langle omega I_n rangle$.






share|cite|improve this answer









$endgroup$



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 $Z(G) = [G,G] = langle omega I_n rangle$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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