How to prove that semidirect product of $Z_{13}$ and $Z_3$ is non Abelian for a non-trivial homomorphism












2












$begingroup$


The semidirect product of $Z_{13}$ and $Z_3$ is given here Finding presentation of group of order 39
as ${x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3}$. I understand how this is arrived at but to show that this group is non abelian, if I take $(x_1,y_1)$ and $(x_2,y_2)$ and look at
$(x_1,y_1).(x_2,y_2) = (x_1 phi_{y_1} (x_2),y_1y_2) = (x_1y_1x_2y_1^{-1},y_1y_2) = (x_1x_2^3,y_1y_2)$
$(x_2,y_2).(x_1,y_1) = (x_2 phi_{y_2} (x_1),y_2y_1) = (x_2y_2x_1y_2^{-1},y_2y_1) = (x_2x_1^3,y_2y_1)$
$y_1y_2 = y_2y_1$ because $y_1,y_2 in Z_3$
$x_1x_2^{3}neq x_2x_1^3 iff x_2x_1x_1^2 neq x_1x_2x_2^2 iff x_1neq x_2 because x_1x_2 = x_2x_1 because x_1,x_2 in Z_{13}$

But this means $(x_1,y_1)$ and $(x_1,y_2)$ commute? I am doing something simple wrong here. Any help is appreciated. Thanks.






abstract-algebra group-theory finite-groups abelian-groups semidirect-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
    $endgroup$
    – Berci
    Dec 2 '18 at 21:54












  • $begingroup$
    @Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:01






  • 1




    $begingroup$
    A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
    $endgroup$
    – Servaes
    Dec 2 '18 at 22:02


















2












$begingroup$


The semidirect product of $Z_{13}$ and $Z_3$ is given here Finding presentation of group of order 39
as ${x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3}$. I understand how this is arrived at but to show that this group is non abelian, if I take $(x_1,y_1)$ and $(x_2,y_2)$ and look at
$(x_1,y_1).(x_2,y_2) = (x_1 phi_{y_1} (x_2),y_1y_2) = (x_1y_1x_2y_1^{-1},y_1y_2) = (x_1x_2^3,y_1y_2)$
$(x_2,y_2).(x_1,y_1) = (x_2 phi_{y_2} (x_1),y_2y_1) = (x_2y_2x_1y_2^{-1},y_2y_1) = (x_2x_1^3,y_2y_1)$
$y_1y_2 = y_2y_1$ because $y_1,y_2 in Z_3$
$x_1x_2^{3}neq x_2x_1^3 iff x_2x_1x_1^2 neq x_1x_2x_2^2 iff x_1neq x_2 because x_1x_2 = x_2x_1 because x_1,x_2 in Z_{13}$

But this means $(x_1,y_1)$ and $(x_1,y_2)$ commute? I am doing something simple wrong here. Any help is appreciated. Thanks.






abstract-algebra group-theory finite-groups abelian-groups semidirect-product







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
    $endgroup$
    – Berci
    Dec 2 '18 at 21:54












  • $begingroup$
    @Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:01






  • 1




    $begingroup$
    A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
    $endgroup$
    – Servaes
    Dec 2 '18 at 22:02
















2












2








2





$begingroup$


The semidirect product of $Z_{13}$ and $Z_3$ is given here Finding presentation of group of order 39
as ${x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3}$. I understand how this is arrived at but to show that this group is non abelian, if I take $(x_1,y_1)$ and $(x_2,y_2)$ and look at
$(x_1,y_1).(x_2,y_2) = (x_1 phi_{y_1} (x_2),y_1y_2) = (x_1y_1x_2y_1^{-1},y_1y_2) = (x_1x_2^3,y_1y_2)$
$(x_2,y_2).(x_1,y_1) = (x_2 phi_{y_2} (x_1),y_2y_1) = (x_2y_2x_1y_2^{-1},y_2y_1) = (x_2x_1^3,y_2y_1)$
$y_1y_2 = y_2y_1$ because $y_1,y_2 in Z_3$
$x_1x_2^{3}neq x_2x_1^3 iff x_2x_1x_1^2 neq x_1x_2x_2^2 iff x_1neq x_2 because x_1x_2 = x_2x_1 because x_1,x_2 in Z_{13}$

But this means $(x_1,y_1)$ and $(x_1,y_2)$ commute? I am doing something simple wrong here. Any help is appreciated. Thanks.






abstract-algebra group-theory finite-groups abelian-groups semidirect-product







share|cite|improve this question











$endgroup$




The semidirect product of $Z_{13}$ and $Z_3$ is given here Finding presentation of group of order 39
as ${x,y | x^{13} = y^3 = 1, yxy^{-1} = x^3}$. I understand how this is arrived at but to show that this group is non abelian, if I take $(x_1,y_1)$ and $(x_2,y_2)$ and look at
$(x_1,y_1).(x_2,y_2) = (x_1 phi_{y_1} (x_2),y_1y_2) = (x_1y_1x_2y_1^{-1},y_1y_2) = (x_1x_2^3,y_1y_2)$
$(x_2,y_2).(x_1,y_1) = (x_2 phi_{y_2} (x_1),y_2y_1) = (x_2y_2x_1y_2^{-1},y_2y_1) = (x_2x_1^3,y_2y_1)$
$y_1y_2 = y_2y_1$ because $y_1,y_2 in Z_3$
$x_1x_2^{3}neq x_2x_1^3 iff x_2x_1x_1^2 neq x_1x_2x_2^2 iff x_1neq x_2 because x_1x_2 = x_2x_1 because x_1,x_2 in Z_{13}$

But this means $(x_1,y_1)$ and $(x_1,y_2)$ commute? I am doing something simple wrong here. Any help is appreciated. Thanks.






abstract-algebra group-theory finite-groups abelian-groups semidirect-product




abstract-algebra group-theory finite-groups abelian-groups semidirect-product






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 22:38









Servaes

25.6k33996




25.6k33996










asked Dec 2 '18 at 21:46









manifoldedmanifolded

1627




1627












  • $begingroup$
    No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
    $endgroup$
    – Berci
    Dec 2 '18 at 21:54












  • $begingroup$
    @Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:01






  • 1




    $begingroup$
    A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
    $endgroup$
    – Servaes
    Dec 2 '18 at 22:02




















  • $begingroup$
    No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
    $endgroup$
    – Berci
    Dec 2 '18 at 21:54












  • $begingroup$
    @Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:01






  • 1




    $begingroup$
    A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
    $endgroup$
    – Servaes
    Dec 2 '18 at 22:02


















$begingroup$
No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
$endgroup$
– Berci
Dec 2 '18 at 21:54






$begingroup$
No, it's all correct. If $x_1ne x_2$, then $(x_1, y_1)$ and $(x_2,y_2)$ don't commute, hence the group is not commutative.
$endgroup$
– Berci
Dec 2 '18 at 21:54














$begingroup$
@Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
$endgroup$
– manifolded
Dec 2 '18 at 22:01




$begingroup$
@Berci But if $x_1 = x_2$ and $y_1 neq y_2$, it should still not commute, right? But it does above.
$endgroup$
– manifolded
Dec 2 '18 at 22:01




1




1




$begingroup$
A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
$endgroup$
– Servaes
Dec 2 '18 at 22:02






$begingroup$
A group is abelian if all pairs of elements commute. A group is not abelian if some pair of elements does not commute.
$endgroup$
– Servaes
Dec 2 '18 at 22:02












1 Answer
1






active

oldest

votes


















2












$begingroup$

Your equivalences show that if $x_1neq x_2$ then $(x_1,y_1)$ and $(x_2,y_2)$ do not commute, whatever $y_1,y_2inBbb{Z}_3$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.



Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $gcdot g^k=g^{k+1}=g^kcdot g$. A concrete example is $S_3$; here we have
$$(1 2)(1 3)neq(1 3)(1 2),$$
so this group is not abelian. But still
$$(1 2 3)(1 3 2)=(1 3 2)(1 2 3).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a ton! Not sure why I forgot that, haha.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:37











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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023254%2fhow-to-prove-that-semidirect-product-of-z-13-and-z-3-is-non-abelian-for-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Your equivalences show that if $x_1neq x_2$ then $(x_1,y_1)$ and $(x_2,y_2)$ do not commute, whatever $y_1,y_2inBbb{Z}_3$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.



Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $gcdot g^k=g^{k+1}=g^kcdot g$. A concrete example is $S_3$; here we have
$$(1 2)(1 3)neq(1 3)(1 2),$$
so this group is not abelian. But still
$$(1 2 3)(1 3 2)=(1 3 2)(1 2 3).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a ton! Not sure why I forgot that, haha.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:37
















2












$begingroup$

Your equivalences show that if $x_1neq x_2$ then $(x_1,y_1)$ and $(x_2,y_2)$ do not commute, whatever $y_1,y_2inBbb{Z}_3$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.



Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $gcdot g^k=g^{k+1}=g^kcdot g$. A concrete example is $S_3$; here we have
$$(1 2)(1 3)neq(1 3)(1 2),$$
so this group is not abelian. But still
$$(1 2 3)(1 3 2)=(1 3 2)(1 2 3).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a ton! Not sure why I forgot that, haha.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:37














2












2








2





$begingroup$

Your equivalences show that if $x_1neq x_2$ then $(x_1,y_1)$ and $(x_2,y_2)$ do not commute, whatever $y_1,y_2inBbb{Z}_3$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.



Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $gcdot g^k=g^{k+1}=g^kcdot g$. A concrete example is $S_3$; here we have
$$(1 2)(1 3)neq(1 3)(1 2),$$
so this group is not abelian. But still
$$(1 2 3)(1 3 2)=(1 3 2)(1 2 3).$$






share|cite|improve this answer









$endgroup$



Your equivalences show that if $x_1neq x_2$ then $(x_1,y_1)$ and $(x_2,y_2)$ do not commute, whatever $y_1,y_2inBbb{Z}_3$ are. So the semidirect product contains elements that do not commute. This means the semidirect product is not abelian.



Note that a group that is not abelian may still contain elements that do commute with eachother. For example, any element in any group commutes with all its own powers, because $gcdot g^k=g^{k+1}=g^kcdot g$. A concrete example is $S_3$; here we have
$$(1 2)(1 3)neq(1 3)(1 2),$$
so this group is not abelian. But still
$$(1 2 3)(1 3 2)=(1 3 2)(1 2 3).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 22:05









ServaesServaes

25.6k33996




25.6k33996












  • $begingroup$
    Thanks a ton! Not sure why I forgot that, haha.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:37


















  • $begingroup$
    Thanks a ton! Not sure why I forgot that, haha.
    $endgroup$
    – manifolded
    Dec 2 '18 at 22:37
















$begingroup$
Thanks a ton! Not sure why I forgot that, haha.
$endgroup$
– manifolded
Dec 2 '18 at 22:37




$begingroup$
Thanks a ton! Not sure why I forgot that, haha.
$endgroup$
– manifolded
Dec 2 '18 at 22:37


















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023254%2fhow-to-prove-that-semidirect-product-of-z-13-and-z-3-is-non-abelian-for-a%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

Image
0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
Read more

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more