Showing $Bbb{Z}_q rtimes Q_8$ has the presentation $langle x,y,z mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1},...












2












$begingroup$


Let $G cong Bbb{Z}_q rtimes Q_8$. Then $G$ has a presentation as follows



$$langle x,y,z mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} rangle.$$




I dont understand why $x^z=x^{-1}$?




(The action is conjugation)






group-theory group-presentation semidirect-product combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
    $endgroup$
    – Lee Mosher
    May 11 '17 at 12:28








  • 1




    $begingroup$
    The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:34








  • 1




    $begingroup$
    There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:36






  • 1




    $begingroup$
    @Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
    $endgroup$
    – Hana
    May 11 '17 at 12:46






  • 2




    $begingroup$
    I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
    $endgroup$
    – Somos
    May 11 '17 at 23:00


















2












$begingroup$


Let $G cong Bbb{Z}_q rtimes Q_8$. Then $G$ has a presentation as follows



$$langle x,y,z mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} rangle.$$




I dont understand why $x^z=x^{-1}$?




(The action is conjugation)






group-theory group-presentation semidirect-product combinatorial-group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
    $endgroup$
    – Lee Mosher
    May 11 '17 at 12:28








  • 1




    $begingroup$
    The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:34








  • 1




    $begingroup$
    There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:36






  • 1




    $begingroup$
    @Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
    $endgroup$
    – Hana
    May 11 '17 at 12:46






  • 2




    $begingroup$
    I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
    $endgroup$
    – Somos
    May 11 '17 at 23:00
















2












2








2


1



$begingroup$


Let $G cong Bbb{Z}_q rtimes Q_8$. Then $G$ has a presentation as follows



$$langle x,y,z mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} rangle.$$




I dont understand why $x^z=x^{-1}$?




(The action is conjugation)






group-theory group-presentation semidirect-product combinatorial-group-theory







share|cite|improve this question











$endgroup$




Let $G cong Bbb{Z}_q rtimes Q_8$. Then $G$ has a presentation as follows



$$langle x,y,z mid x^q=y^4=z^4=[x,y]=1, y^z=y^{-1}, y^2=z^2, x^z=x^{-1} rangle.$$




I dont understand why $x^z=x^{-1}$?




(The action is conjugation)






group-theory group-presentation semidirect-product combinatorial-group-theory




group-theory group-presentation semidirect-product combinatorial-group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 22:40









Shaun

9,083113683




9,083113683










asked May 11 '17 at 12:21









HanaHana

937




937








  • 1




    $begingroup$
    Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
    $endgroup$
    – Lee Mosher
    May 11 '17 at 12:28








  • 1




    $begingroup$
    The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:34








  • 1




    $begingroup$
    There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:36






  • 1




    $begingroup$
    @Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
    $endgroup$
    – Hana
    May 11 '17 at 12:46






  • 2




    $begingroup$
    I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
    $endgroup$
    – Somos
    May 11 '17 at 23:00
















  • 1




    $begingroup$
    Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
    $endgroup$
    – Lee Mosher
    May 11 '17 at 12:28








  • 1




    $begingroup$
    The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:34








  • 1




    $begingroup$
    There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
    $endgroup$
    – Derek Holt
    May 11 '17 at 12:36






  • 1




    $begingroup$
    @Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
    $endgroup$
    – Hana
    May 11 '17 at 12:46






  • 2




    $begingroup$
    I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
    $endgroup$
    – Somos
    May 11 '17 at 23:00










1




1




$begingroup$
Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
$endgroup$
– Lee Mosher
May 11 '17 at 12:28






$begingroup$
Please specify which homomorphism $Q_8 mapsto text{Aut}(mathbb{Z}_q)$ is used to define the semidirect product $mathbb{Z}_q rtimes Q_8;$ until you add that to your question, the semidirect product is not well-defined and your question is unanswerable. Also, once you've specified that homomorphism, it would help if you also added to your question which elements of $mathbb{Z}_q rtimes Q_8$ correspond to the generators $x,y,z$.
$endgroup$
– Lee Mosher
May 11 '17 at 12:28






1




1




$begingroup$
The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
$endgroup$
– Derek Holt
May 11 '17 at 12:34






$begingroup$
The action is conjugation in any semidirect product. To define the group you need to specify a homomorphism from $Q_8$ to the automorphsim group of $C_q$ (the cyclic group of order $q$).
$endgroup$
– Derek Holt
May 11 '17 at 12:34






1




1




$begingroup$
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
$endgroup$
– Derek Holt
May 11 '17 at 12:36




$begingroup$
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
$endgroup$
– Derek Holt
May 11 '17 at 12:36




1




1




$begingroup$
@Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
$endgroup$
– Hana
May 11 '17 at 12:46




$begingroup$
@Derek Holt Ok thats right $y^4=1$. I have a group $G cong C_q rtimes Q_8$. My teacher said its presentation is as above, I dont understand why $x^z=x^{-1}$? I know $x^z in C_q$
$endgroup$
– Hana
May 11 '17 at 12:46




2




2




$begingroup$
I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
$endgroup$
– Somos
May 11 '17 at 23:00






$begingroup$
I am not sure you understand what an abstract grounp presentation is. In your case, there are three generators, $x,y,z$. They generate a free group. Now we require certain relations to hold between the generators. For example $y^4=1$. This gives rise to an equivalence relation of the free group. Similarly, for the other conditions. So $x^z := zxz^{-1} =x^{-1}$ must also hold. When all these relations are factored in, the claim is that the resulting group $G$ is a semi-direct product.
$endgroup$
– Somos
May 11 '17 at 23:00












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