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







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$endgroup$












  • $begingroup$
    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




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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











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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









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


















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