Trouble understanding semidirect product of $Z_3$ and $Z_{13}$












0












$begingroup$


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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 8:08









      Luca Carai

      31119




      31119










      asked Dec 3 '18 at 6:30









      manifoldedmanifolded

      615




      615






















          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 $kin{0,4,8}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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













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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 $kin{0,4,8}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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


















          1












          $begingroup$

          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 $kin{0,4,8}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          1








          1





          $begingroup$

          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 $kin{0,4,8}$.






          share|cite|improve this answer











          $endgroup$



          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 $kin{0,4,8}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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




















          • $begingroup$
            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




















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