In what case the semi direct product is abelian?
$begingroup$
Only when H$rtimes$K is direct product and H,K are abelian? How to prove? In other words if the homomorphic from K to Aut(H) is not trivial then the semi direct product is not abelian?
abstract-algebra group-theory abelian-groups semidirect-product
edited Dec 21 '18 at 16:12
Shaun
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
$endgroup$
add a comment |
$begingroup$
Only when H$rtimes$K is direct product and H,K are abelian? How to prove? In other words if the homomorphic from K to Aut(H) is not trivial then the semi direct product is not abelian?
abstract-algebra group-theory abelian-groups semidirect-product
edited Dec 21 '18 at 16:12
Shaun
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
$endgroup$
add a comment |
$begingroup$
Only when H$rtimes$K is direct product and H,K are abelian? How to prove? In other words if the homomorphic from K to Aut(H) is not trivial then the semi direct product is not abelian?
abstract-algebra group-theory abelian-groups semidirect-product
edited Dec 21 '18 at 16:12
Shaun
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
$endgroup$
Only when H$rtimes$K is direct product and H,K are abelian? How to prove? In other words if the homomorphic from K to Aut(H) is not trivial then the semi direct product is not abelian?
abstract-algebra group-theory abelian-groups semidirect-product
abstract-algebra group-theory abelian-groups semidirect-product
edited Dec 21 '18 at 16:12
Shaun
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
edited Dec 21 '18 at 16:12
Shaun
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
edited Dec 21 '18 at 16:12
Shaun
9,366113684
edited Dec 21 '18 at 16:12
Shaun
9,366113684
edited Dec 21 '18 at 16:12
Shaun
9,366113684
9,366113684
asked Oct 1 '14 at 18:05
66666666
1,343621
asked Oct 1 '14 at 18:05
66666666
1,343621
asked Oct 1 '14 at 18:05
66666666
1,343621
1,343621
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The comutativity is equivalent to
$$y_1 y_2= y_2 y_1 forall y_iin K$$
$$x_1 phi ( y_1 ) ( x_2 ) = x_2phi ( y_2 ) ( x_1 ) forall x_i in H y_i in K$$
The $K$ must be abelian, from the second equation we take
$$phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 phi ( y_1 ) ( x_2 ) $$
So $phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again
$$x_2 x_1 = x_1 x_2 $$
So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
$endgroup$
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
add a comment |
$begingroup$
If $phi:Kto{rm Aut}(H)$ is not trivial then $phi_k:Hto H$ is nontrivial for some $kin K$, so $phi_k(h)ne h$ for some $hin H$, in which case $khk^{-1}=phi_k(h)ne h~,Rightarrow~ khne hk$ within $Hrtimes K$.
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
$endgroup$
add a comment |
$begingroup$
$H$ and $K$ are subgroups of $Hrtimes K$, so if $Hrtimes K$ then $H$ and $K$ had better be abelian too.
The homomorphism $varphi:Ktooperatorname{Aut}(H)$ becomes conjugation in $Hrtimes_varphi K$. Conjugation is always trivial in abelian groups, so again, if $Hrtimes_varphi K$ is abelian, then $varphi$ had better be trivial.
In fact, the following statement is true: $Hrtimes_varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $varphi$ is the trivial homomorphism.
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The comutativity is equivalent to
$$y_1 y_2= y_2 y_1 forall y_iin K$$
$$x_1 phi ( y_1 ) ( x_2 ) = x_2phi ( y_2 ) ( x_1 ) forall x_i in H y_i in K$$
The $K$ must be abelian, from the second equation we take
$$phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 phi ( y_1 ) ( x_2 ) $$
So $phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again
$$x_2 x_1 = x_1 x_2 $$
So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
$endgroup$
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
add a comment |
$begingroup$
The comutativity is equivalent to
$$y_1 y_2= y_2 y_1 forall y_iin K$$
$$x_1 phi ( y_1 ) ( x_2 ) = x_2phi ( y_2 ) ( x_1 ) forall x_i in H y_i in K$$
The $K$ must be abelian, from the second equation we take
$$phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 phi ( y_1 ) ( x_2 ) $$
So $phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again
$$x_2 x_1 = x_1 x_2 $$
So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
$endgroup$
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
add a comment |
$begingroup$
The comutativity is equivalent to
$$y_1 y_2= y_2 y_1 forall y_iin K$$
$$x_1 phi ( y_1 ) ( x_2 ) = x_2phi ( y_2 ) ( x_1 ) forall x_i in H y_i in K$$
The $K$ must be abelian, from the second equation we take
$$phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 phi ( y_1 ) ( x_2 ) $$
So $phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again
$$x_2 x_1 = x_1 x_2 $$
So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
$endgroup$
The comutativity is equivalent to
$$y_1 y_2= y_2 y_1 forall y_iin K$$
$$x_1 phi ( y_1 ) ( x_2 ) = x_2phi ( y_2 ) ( x_1 ) forall x_i in H y_i in K$$
The $K$ must be abelian, from the second equation we take
$$phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 phi ( y_1 ) ( x_2 ) $$
So $phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again
$$x_2 x_1 = x_1 x_2 $$
So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
answered Oct 1 '14 at 18:14
PenasRaulPenasRaul
1,07448
1,07448
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
add a comment |
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3)
$endgroup$
– 6666
Oct 1 '14 at 18:42
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
$begingroup$
$x_3$ does not depend on $y_3$.
$endgroup$
– PenasRaul
Oct 1 '14 at 20:49
add a comment |
$begingroup$
If $phi:Kto{rm Aut}(H)$ is not trivial then $phi_k:Hto H$ is nontrivial for some $kin K$, so $phi_k(h)ne h$ for some $hin H$, in which case $khk^{-1}=phi_k(h)ne h~,Rightarrow~ khne hk$ within $Hrtimes K$.
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
$endgroup$
add a comment |
$begingroup$
If $phi:Kto{rm Aut}(H)$ is not trivial then $phi_k:Hto H$ is nontrivial for some $kin K$, so $phi_k(h)ne h$ for some $hin H$, in which case $khk^{-1}=phi_k(h)ne h~,Rightarrow~ khne hk$ within $Hrtimes K$.
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
$endgroup$
add a comment |
$begingroup$
If $phi:Kto{rm Aut}(H)$ is not trivial then $phi_k:Hto H$ is nontrivial for some $kin K$, so $phi_k(h)ne h$ for some $hin H$, in which case $khk^{-1}=phi_k(h)ne h~,Rightarrow~ khne hk$ within $Hrtimes K$.
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
$endgroup$
If $phi:Kto{rm Aut}(H)$ is not trivial then $phi_k:Hto H$ is nontrivial for some $kin K$, so $phi_k(h)ne h$ for some $hin H$, in which case $khk^{-1}=phi_k(h)ne h~,Rightarrow~ khne hk$ within $Hrtimes K$.
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
answered Oct 1 '14 at 20:31
whackawhacka
12.1k1734
12.1k1734
add a comment |
add a comment |
$begingroup$
$H$ and $K$ are subgroups of $Hrtimes K$, so if $Hrtimes K$ then $H$ and $K$ had better be abelian too.
The homomorphism $varphi:Ktooperatorname{Aut}(H)$ becomes conjugation in $Hrtimes_varphi K$. Conjugation is always trivial in abelian groups, so again, if $Hrtimes_varphi K$ is abelian, then $varphi$ had better be trivial.
In fact, the following statement is true: $Hrtimes_varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $varphi$ is the trivial homomorphism.
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
$endgroup$
add a comment |
$begingroup$
$H$ and $K$ are subgroups of $Hrtimes K$, so if $Hrtimes K$ then $H$ and $K$ had better be abelian too.
The homomorphism $varphi:Ktooperatorname{Aut}(H)$ becomes conjugation in $Hrtimes_varphi K$. Conjugation is always trivial in abelian groups, so again, if $Hrtimes_varphi K$ is abelian, then $varphi$ had better be trivial.
In fact, the following statement is true: $Hrtimes_varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $varphi$ is the trivial homomorphism.
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
$endgroup$
add a comment |
$begingroup$
$H$ and $K$ are subgroups of $Hrtimes K$, so if $Hrtimes K$ then $H$ and $K$ had better be abelian too.
The homomorphism $varphi:Ktooperatorname{Aut}(H)$ becomes conjugation in $Hrtimes_varphi K$. Conjugation is always trivial in abelian groups, so again, if $Hrtimes_varphi K$ is abelian, then $varphi$ had better be trivial.
In fact, the following statement is true: $Hrtimes_varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $varphi$ is the trivial homomorphism.
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
$endgroup$
$H$ and $K$ are subgroups of $Hrtimes K$, so if $Hrtimes K$ then $H$ and $K$ had better be abelian too.
The homomorphism $varphi:Ktooperatorname{Aut}(H)$ becomes conjugation in $Hrtimes_varphi K$. Conjugation is always trivial in abelian groups, so again, if $Hrtimes_varphi K$ is abelian, then $varphi$ had better be trivial.
In fact, the following statement is true: $Hrtimes_varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $varphi$ is the trivial homomorphism.
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
answered Oct 1 '14 at 20:43
SladeSlade
25.2k12665
25.2k12665
add a comment |
add a comment |
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