Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$?












3












$begingroup$


Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?



In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?



I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?



Thank you so much in advance.






group-theory representation-theory







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?



    In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?



    I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?



    Thank you so much in advance.






    group-theory representation-theory







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?



      In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?



      I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?



      Thank you so much in advance.






      group-theory representation-theory







      share|cite|improve this question











      $endgroup$




      Does alternating group $A_5$ is a subgroup of $GL(4,mathbb Z)$? I know $A_5$ has a 4-dimensional complex representation. But how to prove $A_5$ is a subgroup or not of $GL(4, mathbb Z)$?



      In case $GL(4, mathbb Z)$ has a subgroup isomorphic to $A_5$, then I would like to know corresponding generators of that subgroup?



      I want to know how many subgroups are there in $GL(4, Bbb Z)$ isomorphic to $A_5$. Is there a unique one?



      Thank you so much in advance.






      group-theory representation-theory




      group-theory representation-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 8:06







      mathstudent

















      asked Jan 2 at 16:11









      mathstudentmathstudent

      394




      394






















          1 Answer
          1






          active

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          3












          $begingroup$

          For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.



          Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
          are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
          By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
          a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
          (this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
          Bbb Z^{n-1})
          =GL(n-1,Bbb Z)$          .



          In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
            $endgroup$
            – mathstudent
            Jan 2 at 17:00












          • $begingroup$
            What are your favourite generators of $A_5$? @mathstudent
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 17:02










          • $begingroup$
            Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
            $endgroup$
            – mathstudent
            Jan 2 at 17:04








          • 1




            $begingroup$
            It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
            $endgroup$
            – Derek Holt
            Jan 3 at 8:28








          • 1




            $begingroup$
            I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
            $endgroup$
            – Derek Holt
            Jan 3 at 9:08













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          3












          $begingroup$

          For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.



          Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
          are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
          By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
          a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
          (this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
          Bbb Z^{n-1})
          =GL(n-1,Bbb Z)$          .



          In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
            $endgroup$
            – mathstudent
            Jan 2 at 17:00












          • $begingroup$
            What are your favourite generators of $A_5$? @mathstudent
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 17:02










          • $begingroup$
            Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
            $endgroup$
            – mathstudent
            Jan 2 at 17:04








          • 1




            $begingroup$
            It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
            $endgroup$
            – Derek Holt
            Jan 3 at 8:28








          • 1




            $begingroup$
            I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
            $endgroup$
            – Derek Holt
            Jan 3 at 9:08


















          3












          $begingroup$

          For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.



          Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
          are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
          By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
          a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
          (this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
          Bbb Z^{n-1})
          =GL(n-1,Bbb Z)$          .



          In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
            $endgroup$
            – mathstudent
            Jan 2 at 17:00












          • $begingroup$
            What are your favourite generators of $A_5$? @mathstudent
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 17:02










          • $begingroup$
            Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
            $endgroup$
            – mathstudent
            Jan 2 at 17:04








          • 1




            $begingroup$
            It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
            $endgroup$
            – Derek Holt
            Jan 3 at 8:28








          • 1




            $begingroup$
            I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
            $endgroup$
            – Derek Holt
            Jan 3 at 9:08
















          3












          3








          3





          $begingroup$

          For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.



          Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
          are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
          By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
          a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
          (this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
          Bbb Z^{n-1})
          =GL(n-1,Bbb Z)$          .



          In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.






          share|cite|improve this answer









          $endgroup$



          For any $n$, $S_n$ embeds in $GL(n-1,Bbb Z)$. Therefore $A_n$ does too.



          Consider the vectors $v_1,ldots,v_n$ where $v_1,ldots,v_{n-1}$
          are the standard unit vectors in $Bbb Z^{n-1}$ and $v_n=-v_1-cdots-v_{n-1}$.
          By definition $v_1+cdots+v_n=0$. For each permutation $tauin S_n$, there's
          a unique endomorphism of $Bbb Z^{n-1}$ taking each $v_k$ to $v_{tau(k)}$
          (this works due to $v_1+cdots+v_n=0$). So $S_n$ embeds into $text{Aut}(
          Bbb Z^{n-1})
          =GL(n-1,Bbb Z)$          .



          In fact $A_n$ will embed in $SL(n-1,Bbb Z)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 16:17









          Lord Shark the UnknownLord Shark the Unknown

          105k1160133




          105k1160133












          • $begingroup$
            Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
            $endgroup$
            – mathstudent
            Jan 2 at 17:00












          • $begingroup$
            What are your favourite generators of $A_5$? @mathstudent
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 17:02










          • $begingroup$
            Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
            $endgroup$
            – mathstudent
            Jan 2 at 17:04








          • 1




            $begingroup$
            It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
            $endgroup$
            – Derek Holt
            Jan 3 at 8:28








          • 1




            $begingroup$
            I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
            $endgroup$
            – Derek Holt
            Jan 3 at 9:08




















          • $begingroup$
            Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
            $endgroup$
            – mathstudent
            Jan 2 at 17:00












          • $begingroup$
            What are your favourite generators of $A_5$? @mathstudent
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 17:02










          • $begingroup$
            Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
            $endgroup$
            – mathstudent
            Jan 2 at 17:04








          • 1




            $begingroup$
            It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
            $endgroup$
            – Derek Holt
            Jan 3 at 8:28








          • 1




            $begingroup$
            I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
            $endgroup$
            – Derek Holt
            Jan 3 at 9:08


















          $begingroup$
          Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
          $endgroup$
          – mathstudent
          Jan 2 at 17:00






          $begingroup$
          Thanks. I would like to know corresponding generators of the subgroup of $GL(4, mathbb Z)$ that is isomorphic to $A_5$ if possible.
          $endgroup$
          – mathstudent
          Jan 2 at 17:00














          $begingroup$
          What are your favourite generators of $A_5$? @mathstudent
          $endgroup$
          – Lord Shark the Unknown
          Jan 2 at 17:02




          $begingroup$
          What are your favourite generators of $A_5$? @mathstudent
          $endgroup$
          – Lord Shark the Unknown
          Jan 2 at 17:02












          $begingroup$
          Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
          $endgroup$
          – mathstudent
          Jan 2 at 17:04






          $begingroup$
          Generators of $A_5$: $<(1 2)(3 4), (1 2 3 4 5)>$. Thanks.
          $endgroup$
          – mathstudent
          Jan 2 at 17:04






          1




          1




          $begingroup$
          It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
          $endgroup$
          – Derek Holt
          Jan 3 at 8:28






          $begingroup$
          It is a bad idea to ask new questions in comments. It follows from the representation theory over fields that the subgroup is unique up to conjugacy in ${rm GL}(4,{mathbb Q})$. In general, conjugacy in ${rm GL}(4,{mathbb Z})$ is more difficult to determine. I don't know the answer in this case, but it is certainly known, because the finite subgroups of ${rm GL}(n,{mathbb Z})$ have been classified up to about $n=24$.
          $endgroup$
          – Derek Holt
          Jan 3 at 8:28






          1




          1




          $begingroup$
          I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
          $endgroup$
          – Derek Holt
          Jan 3 at 9:08






          $begingroup$
          I just did a calculation in Magma. There are two ${rm GL}(4,{mathbb Z})$-classes of subgroups isomorphic to $A_5$, which are conjugate in ${rm GL}(4,{mathbb Q})$. You can get one class from the other by taking the dual i.e. inverting and transposing the matrices. By the way, the third column of your matrix for $(1,2,5,3,4)$ is wrong - it should by $(0,0,0,1)$.
          $endgroup$
          – Derek Holt
          Jan 3 at 9:08




















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