Trouble understanding semidirect product of $Z_3$ and $Z_{13}$
$begingroup$
Here a representation for a non Abelian of order 39 is given
Finding presentation of group of order 39 by Pisco. Am I correct in understanding that the non-trivial homomorphism in this case is:
$phi:Z_3 rightarrow Aut(Z_{13})$ where $phi_y(x) = yxy^{-1}$ $forall yin Z_3$ and $forall xin Z_{13}$
For different $y$ we get different Automorphisms in $Z_{13}$.
To find the actual automorphism i.e. to find $i$ s.t. $yxy^{-1} = x^i$, we are using the automorphism corresponding to the generator $y$ of $Z_3$ and $x$ of $Z_{13}$? Appreciate your help in explaining the semidirect product in terms of its explicit definition to help me get used to the idea.
abstract-algebra group-theory finite-groups
$endgroup$
add a comment |
$begingroup$
Here a representation for a non Abelian of order 39 is given
Finding presentation of group of order 39 by Pisco. Am I correct in understanding that the non-trivial homomorphism in this case is:
$phi:Z_3 rightarrow Aut(Z_{13})$ where $phi_y(x) = yxy^{-1}$ $forall yin Z_3$ and $forall xin Z_{13}$
For different $y$ we get different Automorphisms in $Z_{13}$.
To find the actual automorphism i.e. to find $i$ s.t. $yxy^{-1} = x^i$, we are using the automorphism corresponding to the generator $y$ of $Z_3$ and $x$ of $Z_{13}$? Appreciate your help in explaining the semidirect product in terms of its explicit definition to help me get used to the idea.
abstract-algebra group-theory finite-groups
$endgroup$
add a comment |
$begingroup$
Here a representation for a non Abelian of order 39 is given
Finding presentation of group of order 39 by Pisco. Am I correct in understanding that the non-trivial homomorphism in this case is:
$phi:Z_3 rightarrow Aut(Z_{13})$ where $phi_y(x) = yxy^{-1}$ $forall yin Z_3$ and $forall xin Z_{13}$
For different $y$ we get different Automorphisms in $Z_{13}$.
To find the actual automorphism i.e. to find $i$ s.t. $yxy^{-1} = x^i$, we are using the automorphism corresponding to the generator $y$ of $Z_3$ and $x$ of $Z_{13}$? Appreciate your help in explaining the semidirect product in terms of its explicit definition to help me get used to the idea.
abstract-algebra group-theory finite-groups
$endgroup$
Here a representation for a non Abelian of order 39 is given
Finding presentation of group of order 39 by Pisco. Am I correct in understanding that the non-trivial homomorphism in this case is:
$phi:Z_3 rightarrow Aut(Z_{13})$ where $phi_y(x) = yxy^{-1}$ $forall yin Z_3$ and $forall xin Z_{13}$
For different $y$ we get different Automorphisms in $Z_{13}$.
To find the actual automorphism i.e. to find $i$ s.t. $yxy^{-1} = x^i$, we are using the automorphism corresponding to the generator $y$ of $Z_3$ and $x$ of $Z_{13}$? Appreciate your help in explaining the semidirect product in terms of its explicit definition to help me get used to the idea.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Dec 3 '18 at 8:08
Luca Carai
31119
31119
asked Dec 3 '18 at 6:30
manifoldedmanifolded
615
615
add a comment |
add a comment |
1 Answer
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$begingroup$
Your homomorphism $phi$ makes no sense to me; in
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): y longmapsto (x longmapsto yxy^{-1}),$$
how do you multiply elements $yinBbb{Z}/3Bbb{Z}$ and $xinBbb{Z}/13Bbb{Z}$ together? Let me give some background.
In stead, identify $operatorname{Aut}(Bbb{Z}/13Bbb{Z})cong(Bbb{Z}/13Bbb{Z})^{times}$ with $Bbb{Z}/12Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(Bbb{Z}/13Bbb{Z})^{times}$.
For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $Bbb{Z}/13Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form
$$psi: Bbb{Z}/13Bbb{Z} longmapsto Bbb{Z}/13Bbb{Z}: 2 longmapsto 2^k,$$
for some $kin{1,ldots,12}$. Check that this gives an isomorphism
$$Bbb{Z}/12Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): k longmapsto (n longmapsto ncdot2^k).$$
This makes it clearer what homomorphisms $phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z})$ look like. They are determined by where $1inBbb{Z}/3Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $Bbb{Z}/12Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}),$$
and they are of the form
$$phi(m): Bbb{Z}/13Bbb{Z} longrightarrow Bbb{Z}/13Bbb{Z}: n longmapsto ncdot2^{km},$$
for some $kin{0,4,8}$.
$endgroup$
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your homomorphism $phi$ makes no sense to me; in
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): y longmapsto (x longmapsto yxy^{-1}),$$
how do you multiply elements $yinBbb{Z}/3Bbb{Z}$ and $xinBbb{Z}/13Bbb{Z}$ together? Let me give some background.
In stead, identify $operatorname{Aut}(Bbb{Z}/13Bbb{Z})cong(Bbb{Z}/13Bbb{Z})^{times}$ with $Bbb{Z}/12Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(Bbb{Z}/13Bbb{Z})^{times}$.
For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $Bbb{Z}/13Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form
$$psi: Bbb{Z}/13Bbb{Z} longmapsto Bbb{Z}/13Bbb{Z}: 2 longmapsto 2^k,$$
for some $kin{1,ldots,12}$. Check that this gives an isomorphism
$$Bbb{Z}/12Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): k longmapsto (n longmapsto ncdot2^k).$$
This makes it clearer what homomorphisms $phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z})$ look like. They are determined by where $1inBbb{Z}/3Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $Bbb{Z}/12Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}),$$
and they are of the form
$$phi(m): Bbb{Z}/13Bbb{Z} longrightarrow Bbb{Z}/13Bbb{Z}: n longmapsto ncdot2^{km},$$
for some $kin{0,4,8}$.
$endgroup$
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
add a comment |
$begingroup$
Your homomorphism $phi$ makes no sense to me; in
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): y longmapsto (x longmapsto yxy^{-1}),$$
how do you multiply elements $yinBbb{Z}/3Bbb{Z}$ and $xinBbb{Z}/13Bbb{Z}$ together? Let me give some background.
In stead, identify $operatorname{Aut}(Bbb{Z}/13Bbb{Z})cong(Bbb{Z}/13Bbb{Z})^{times}$ with $Bbb{Z}/12Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(Bbb{Z}/13Bbb{Z})^{times}$.
For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $Bbb{Z}/13Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form
$$psi: Bbb{Z}/13Bbb{Z} longmapsto Bbb{Z}/13Bbb{Z}: 2 longmapsto 2^k,$$
for some $kin{1,ldots,12}$. Check that this gives an isomorphism
$$Bbb{Z}/12Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): k longmapsto (n longmapsto ncdot2^k).$$
This makes it clearer what homomorphisms $phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z})$ look like. They are determined by where $1inBbb{Z}/3Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $Bbb{Z}/12Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}),$$
and they are of the form
$$phi(m): Bbb{Z}/13Bbb{Z} longrightarrow Bbb{Z}/13Bbb{Z}: n longmapsto ncdot2^{km},$$
for some $kin{0,4,8}$.
$endgroup$
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
add a comment |
$begingroup$
Your homomorphism $phi$ makes no sense to me; in
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): y longmapsto (x longmapsto yxy^{-1}),$$
how do you multiply elements $yinBbb{Z}/3Bbb{Z}$ and $xinBbb{Z}/13Bbb{Z}$ together? Let me give some background.
In stead, identify $operatorname{Aut}(Bbb{Z}/13Bbb{Z})cong(Bbb{Z}/13Bbb{Z})^{times}$ with $Bbb{Z}/12Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(Bbb{Z}/13Bbb{Z})^{times}$.
For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $Bbb{Z}/13Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form
$$psi: Bbb{Z}/13Bbb{Z} longmapsto Bbb{Z}/13Bbb{Z}: 2 longmapsto 2^k,$$
for some $kin{1,ldots,12}$. Check that this gives an isomorphism
$$Bbb{Z}/12Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): k longmapsto (n longmapsto ncdot2^k).$$
This makes it clearer what homomorphisms $phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z})$ look like. They are determined by where $1inBbb{Z}/3Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $Bbb{Z}/12Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}),$$
and they are of the form
$$phi(m): Bbb{Z}/13Bbb{Z} longrightarrow Bbb{Z}/13Bbb{Z}: n longmapsto ncdot2^{km},$$
for some $kin{0,4,8}$.
$endgroup$
Your homomorphism $phi$ makes no sense to me; in
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): y longmapsto (x longmapsto yxy^{-1}),$$
how do you multiply elements $yinBbb{Z}/3Bbb{Z}$ and $xinBbb{Z}/13Bbb{Z}$ together? Let me give some background.
In stead, identify $operatorname{Aut}(Bbb{Z}/13Bbb{Z})cong(Bbb{Z}/13Bbb{Z})^{times}$ with $Bbb{Z}/12Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(Bbb{Z}/13Bbb{Z})^{times}$.
For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $Bbb{Z}/13Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form
$$psi: Bbb{Z}/13Bbb{Z} longmapsto Bbb{Z}/13Bbb{Z}: 2 longmapsto 2^k,$$
for some $kin{1,ldots,12}$. Check that this gives an isomorphism
$$Bbb{Z}/12Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}): k longmapsto (n longmapsto ncdot2^k).$$
This makes it clearer what homomorphisms $phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z})$ look like. They are determined by where $1inBbb{Z}/3Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $Bbb{Z}/12Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms
$$phi: Bbb{Z}/3Bbb{Z} longrightarrow operatorname{Aut}(Bbb{Z}/13Bbb{Z}),$$
and they are of the form
$$phi(m): Bbb{Z}/13Bbb{Z} longrightarrow Bbb{Z}/13Bbb{Z}: n longmapsto ncdot2^{km},$$
for some $kin{0,4,8}$.
edited Dec 4 '18 at 8:07
answered Dec 3 '18 at 8:34
ServaesServaes
23.4k33893
23.4k33893
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
add a comment |
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
$begingroup$
Here $m in Z/3Z$ right? So, for all $m$, does $phi$ map $m$ to $n rightarrow n.2^k$?
$endgroup$
– manifolded
Dec 3 '18 at 22:56
1
1
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
I can't understand how $phi$ is not a function of $m$.
$endgroup$
– manifolded
Dec 3 '18 at 23:00
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
$begingroup$
@manifolded Oops, I forgot an exponent. I've corrected it now.
$endgroup$
– Servaes
Dec 4 '18 at 8:06
add a comment |
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