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
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
edited Nov 29 '18 at 22:40
Shaun
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
|
show 2 more comments
2
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
edited Nov 29 '18 at 22:40
Shaun
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
1
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
1
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 $q$).
– Derek Holt
May 11 '17 at 12:34
1
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
1
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
2
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.
– Somos
May 11 '17 at 23:00
|
show 2 more comments
2
2
2
1
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
edited Nov 29 '18 at 22:40
Shaun
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
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
edited Nov 29 '18 at 22:40
Shaun
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
edited Nov 29 '18 at 22:40
Shaun
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
edited Nov 29 '18 at 22:40
Shaun
8,820113681
edited Nov 29 '18 at 22:40
Shaun
8,820113681
edited Nov 29 '18 at 22:40
Shaun
8,820113681
8,820113681
asked May 11 '17 at 12:21
HanaHana
937
asked May 11 '17 at 12:21
HanaHana
937
asked May 11 '17 at 12:21
HanaHana
937
937
1
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
1
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 $q$).
– Derek Holt
May 11 '17 at 12:34
1
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
1
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
2
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.
– Somos
May 11 '17 at 23:00
|
show 2 more comments
1
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
1
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 $q$).
– Derek Holt
May 11 '17 at 12:34
1
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
1
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
2
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.
– Somos
May 11 '17 at 23:00
1
1
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
1
1
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 $q$).
– Derek Holt
May 11 '17 at 12:34
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 $q$).
– Derek Holt
May 11 '17 at 12:34
1
1
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
1
1
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
2
2
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.
– Somos
May 11 '17 at 23:00
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.
– Somos
May 11 '17 at 23:00
|
show 2 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2276275%2fshowing-bbbz-q-rtimes-q-8-has-the-presentation-langle-x-y-z-mid-xq-y4%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2276275%2fshowing-bbbz-q-rtimes-q-8-has-the-presentation-langle-x-y-z-mid-xq-y4%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
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 $x,y,z$.
– Lee Mosher
May 11 '17 at 12:28
1
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 $q$).
– Derek Holt
May 11 '17 at 12:34
1
There is also a typo in your presentation. It should be $y^4=1$, not $y^2=1$.
– Derek Holt
May 11 '17 at 12:36
1
@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 $x^z in C_q$
– Hana
May 11 '17 at 12:46
2
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.
– Somos
May 11 '17 at 23:00