Subgroup of general linear group over $mathbb{F}_{3}$ generated by two matrices












1












$begingroup$


I have two matrices: $A=begin{bmatrix}0 & 2 &1\1 & 2&1\1 &2&0end{bmatrix}$
and $B=begin{bmatrix}2 & 1 &0\0 & 1&0\0 &1&2end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, say G, is isomorphic to $Q_{8}$ or $D_{4}$ or $S_{3}$ or $S_{4}$ or $A_{4}$. Any hints?






abstract-algebra group-theory







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$endgroup$








  • 1




    $begingroup$
    What is your work on the subject ?
    $endgroup$
    – Jean Marie
    Feb 2 at 21:19










  • $begingroup$
    @JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
    $endgroup$
    – Chyma
    Feb 2 at 21:20










  • $begingroup$
    @JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
    $endgroup$
    – Chyma
    Feb 2 at 21:46












  • $begingroup$
    Are you sure $BABneq A^3$? I get that it does on Mathematica.
    $endgroup$
    – Alec B-G
    Feb 2 at 22:00












  • $begingroup$
    @AlecB-G oops. Thank you!
    $endgroup$
    – Chyma
    Feb 2 at 22:06
















1












$begingroup$


I have two matrices: $A=begin{bmatrix}0 & 2 &1\1 & 2&1\1 &2&0end{bmatrix}$
and $B=begin{bmatrix}2 & 1 &0\0 & 1&0\0 &1&2end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, say G, is isomorphic to $Q_{8}$ or $D_{4}$ or $S_{3}$ or $S_{4}$ or $A_{4}$. Any hints?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is your work on the subject ?
    $endgroup$
    – Jean Marie
    Feb 2 at 21:19










  • $begingroup$
    @JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
    $endgroup$
    – Chyma
    Feb 2 at 21:20










  • $begingroup$
    @JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
    $endgroup$
    – Chyma
    Feb 2 at 21:46












  • $begingroup$
    Are you sure $BABneq A^3$? I get that it does on Mathematica.
    $endgroup$
    – Alec B-G
    Feb 2 at 22:00












  • $begingroup$
    @AlecB-G oops. Thank you!
    $endgroup$
    – Chyma
    Feb 2 at 22:06














1












1








1





$begingroup$


I have two matrices: $A=begin{bmatrix}0 & 2 &1\1 & 2&1\1 &2&0end{bmatrix}$
and $B=begin{bmatrix}2 & 1 &0\0 & 1&0\0 &1&2end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, say G, is isomorphic to $Q_{8}$ or $D_{4}$ or $S_{3}$ or $S_{4}$ or $A_{4}$. Any hints?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




I have two matrices: $A=begin{bmatrix}0 & 2 &1\1 & 2&1\1 &2&0end{bmatrix}$
and $B=begin{bmatrix}2 & 1 &0\0 & 1&0\0 &1&2end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, say G, is isomorphic to $Q_{8}$ or $D_{4}$ or $S_{3}$ or $S_{4}$ or $A_{4}$. Any hints?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 at 0:48









Sam Hughes

743114




743114










asked Feb 2 at 21:10









ChymaChyma

143




143








  • 1




    $begingroup$
    What is your work on the subject ?
    $endgroup$
    – Jean Marie
    Feb 2 at 21:19










  • $begingroup$
    @JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
    $endgroup$
    – Chyma
    Feb 2 at 21:20










  • $begingroup$
    @JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
    $endgroup$
    – Chyma
    Feb 2 at 21:46












  • $begingroup$
    Are you sure $BABneq A^3$? I get that it does on Mathematica.
    $endgroup$
    – Alec B-G
    Feb 2 at 22:00












  • $begingroup$
    @AlecB-G oops. Thank you!
    $endgroup$
    – Chyma
    Feb 2 at 22:06














  • 1




    $begingroup$
    What is your work on the subject ?
    $endgroup$
    – Jean Marie
    Feb 2 at 21:19










  • $begingroup$
    @JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
    $endgroup$
    – Chyma
    Feb 2 at 21:20










  • $begingroup$
    @JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
    $endgroup$
    – Chyma
    Feb 2 at 21:46












  • $begingroup$
    Are you sure $BABneq A^3$? I get that it does on Mathematica.
    $endgroup$
    – Alec B-G
    Feb 2 at 22:00












  • $begingroup$
    @AlecB-G oops. Thank you!
    $endgroup$
    – Chyma
    Feb 2 at 22:06








1




1




$begingroup$
What is your work on the subject ?
$endgroup$
– Jean Marie
Feb 2 at 21:19




$begingroup$
What is your work on the subject ?
$endgroup$
– Jean Marie
Feb 2 at 21:19












$begingroup$
@JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
$endgroup$
– Chyma
Feb 2 at 21:20




$begingroup$
@JeanMarie i found that $A^4 = 1$ and $B^2 = 1$
$endgroup$
– Chyma
Feb 2 at 21:20












$begingroup$
@JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
$endgroup$
– Chyma
Feb 2 at 21:46






$begingroup$
@JeanMarie i dont understand. I tried to show that G isomorphic to $D_{4}$ but $BAB = A^{-1}$ fails ( $D_{4} = <x,y|x^4 = y^2 = 1, yxy = x^{-1}>$ if i'm right)
$endgroup$
– Chyma
Feb 2 at 21:46














$begingroup$
Are you sure $BABneq A^3$? I get that it does on Mathematica.
$endgroup$
– Alec B-G
Feb 2 at 22:00






$begingroup$
Are you sure $BABneq A^3$? I get that it does on Mathematica.
$endgroup$
– Alec B-G
Feb 2 at 22:00














$begingroup$
@AlecB-G oops. Thank you!
$endgroup$
– Chyma
Feb 2 at 22:06




$begingroup$
@AlecB-G oops. Thank you!
$endgroup$
– Chyma
Feb 2 at 22:06










1 Answer
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$begingroup$

We have $A^2=begin{bmatrix}0&0&2\ 0&2&0\ 2&0&0end{bmatrix}$, $A^4=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$, $B^2=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$ and $BAB=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$.



The dihedral group $D_8$ has presentation $langle a,b | a^4=b^2=1, bab=a^{-1}rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $langle A,Branglecong D_8$.






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    $begingroup$

    We have $A^2=begin{bmatrix}0&0&2\ 0&2&0\ 2&0&0end{bmatrix}$, $A^4=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$, $B^2=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$ and $BAB=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$.



    The dihedral group $D_8$ has presentation $langle a,b | a^4=b^2=1, bab=a^{-1}rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $langle A,Branglecong D_8$.






    share|cite|improve this answer









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      $begingroup$

      We have $A^2=begin{bmatrix}0&0&2\ 0&2&0\ 2&0&0end{bmatrix}$, $A^4=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$, $B^2=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$ and $BAB=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$.



      The dihedral group $D_8$ has presentation $langle a,b | a^4=b^2=1, bab=a^{-1}rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $langle A,Branglecong D_8$.






      share|cite|improve this answer









      $endgroup$
















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        $begingroup$

        We have $A^2=begin{bmatrix}0&0&2\ 0&2&0\ 2&0&0end{bmatrix}$, $A^4=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$, $B^2=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$ and $BAB=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$.



        The dihedral group $D_8$ has presentation $langle a,b | a^4=b^2=1, bab=a^{-1}rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $langle A,Branglecong D_8$.






        share|cite|improve this answer









        $endgroup$



        We have $A^2=begin{bmatrix}0&0&2\ 0&2&0\ 2&0&0end{bmatrix}$, $A^4=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$, $B^2=begin{bmatrix}1&0&0\ 0&1&0\ 0&0&1end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$ and $BAB=begin{bmatrix}2&1&0\ 2&1&2\ 0&1&2end{bmatrix}$.



        The dihedral group $D_8$ has presentation $langle a,b | a^4=b^2=1, bab=a^{-1}rangle$, clearly $A$ and $B$ satisfy these relations, so we can conclude $langle A,Branglecong D_8$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 2 at 23:44









        Sam HughesSam Hughes

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