$(mathbb Z/p mathbb Z rtimes mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z congmathbb Z/p mathbb Z rtimes...











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4
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Given:



Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.



It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.



Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.



Note that composition of maps is evaluated from right to left.



Question:



If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?



Thoughts:



My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.



Makes this line of attack sense? Or is the required argument quite obvious?



Thank you very much for your insights!



Context:



I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.










share|cite|improve this question


























    up vote
    4
    down vote

    favorite












    Given:



    Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.



    It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.



    Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.



    Note that composition of maps is evaluated from right to left.



    Question:



    If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?



    Thoughts:



    My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.



    Makes this line of attack sense? Or is the required argument quite obvious?



    Thank you very much for your insights!



    Context:



    I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.










    share|cite|improve this question
























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      Given:



      Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.



      It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.



      Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.



      Note that composition of maps is evaluated from right to left.



      Question:



      If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?



      Thoughts:



      My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.



      Makes this line of attack sense? Or is the required argument quite obvious?



      Thank you very much for your insights!



      Context:



      I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.










      share|cite|improve this question













      Given:



      Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.



      It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.



      Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.



      Note that composition of maps is evaluated from right to left.



      Question:



      If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?



      Thoughts:



      My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.



      Makes this line of attack sense? Or is the required argument quite obvious?



      Thank you very much for your insights!



      Context:



      I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.







      group-theory finite-groups group-isomorphism semidirect-product






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      asked Nov 17 at 20:10









      Moritz

      1,1511622




      1,1511622






















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          I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
          The second way is to show that $H$ "has the same structure as $G$".



          Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!






          share|cite|improve this answer





















          • Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
            – Moritz
            Nov 18 at 15:04











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
          The second way is to show that $H$ "has the same structure as $G$".



          Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!






          share|cite|improve this answer





















          • Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
            – Moritz
            Nov 18 at 15:04















          up vote
          3
          down vote



          accepted










          I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
          The second way is to show that $H$ "has the same structure as $G$".



          Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!






          share|cite|improve this answer





















          • Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
            – Moritz
            Nov 18 at 15:04













          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
          The second way is to show that $H$ "has the same structure as $G$".



          Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!






          share|cite|improve this answer












          I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
          The second way is to show that $H$ "has the same structure as $G$".



          Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 21:30









          Pietro Gheri

          1362




          1362












          • Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
            – Moritz
            Nov 18 at 15:04


















          • Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
            – Moritz
            Nov 18 at 15:04
















          Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
          – Moritz
          Nov 18 at 15:04




          Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
          – Moritz
          Nov 18 at 15:04


















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