subgroup of $GL(2,mathbb{R})$, set of all invertible matrices over $mathbb{R}$












-5












$begingroup$


Does $GL(2,mathbb{R})$ contain a cyclic subgroup of order $5$, where $mathbb{R}$ = set of all real numbers?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$

















    -5












    $begingroup$


    Does $GL(2,mathbb{R})$ contain a cyclic subgroup of order $5$, where $mathbb{R}$ = set of all real numbers?






    abstract-algebra group-theory







    share|cite|improve this question











    $endgroup$















      -5












      -5








      -5





      $begingroup$


      Does $GL(2,mathbb{R})$ contain a cyclic subgroup of order $5$, where $mathbb{R}$ = set of all real numbers?






      abstract-algebra group-theory







      share|cite|improve this question











      $endgroup$




      Does $GL(2,mathbb{R})$ contain a cyclic subgroup of order $5$, where $mathbb{R}$ = set of all real numbers?






      abstract-algebra group-theory




      abstract-algebra group-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 21 at 20:35









      user26857

      39.3k124183




      39.3k124183










      asked Jan 21 at 7:47









      choton chotonchoton choton

      11




      11






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Hint: In any group, any element of order $5$ generates a cyclic subgroup of order $5$. Can you think of an invertible linear transformation $T$ of the plane for which $T^5 = I$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            but then entries of the matrix will be complex numbers
            $endgroup$
            – choton choton
            Jan 21 at 7:57






          • 1




            $begingroup$
            Says who?$mathstrut$
            $endgroup$
            – Ivan Neretin
            Jan 21 at 8:04






          • 1




            $begingroup$
            @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
            $endgroup$
            – Arthur
            Jan 21 at 8:10












          • $begingroup$
            actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
            $endgroup$
            – choton choton
            Jan 21 at 10:18












          • $begingroup$
            @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
            $endgroup$
            – Arthur
            Jan 21 at 10:23





















          1












          $begingroup$

          Any cyclic group can be regarded as a subgroup of $mathrm{GL}(2, mathbb{R})$. Note that it is always good to regard matrices as linear transformations. For the finite cyclic groups, try to think about a linear transformation that gives you the identity transformation if you apply it for the finite times, say, $n$ times. (Think about the rotation.) For the infinite cyclic group, consider the following matrix:
          $$
          T=begin{pmatrix} 1& 1\0&1end{pmatrix}.
          $$
          How does this matrix act on the real plane $mathbb{R}^{2}$ and why this matrix has infinite order?






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Hint: In any group, any element of order $5$ generates a cyclic subgroup of order $5$. Can you think of an invertible linear transformation $T$ of the plane for which $T^5 = I$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              but then entries of the matrix will be complex numbers
              $endgroup$
              – choton choton
              Jan 21 at 7:57






            • 1




              $begingroup$
              Says who?$mathstrut$
              $endgroup$
              – Ivan Neretin
              Jan 21 at 8:04






            • 1




              $begingroup$
              @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
              $endgroup$
              – Arthur
              Jan 21 at 8:10












            • $begingroup$
              actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
              $endgroup$
              – choton choton
              Jan 21 at 10:18












            • $begingroup$
              @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
              $endgroup$
              – Arthur
              Jan 21 at 10:23


















            4












            $begingroup$

            Hint: In any group, any element of order $5$ generates a cyclic subgroup of order $5$. Can you think of an invertible linear transformation $T$ of the plane for which $T^5 = I$?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              but then entries of the matrix will be complex numbers
              $endgroup$
              – choton choton
              Jan 21 at 7:57






            • 1




              $begingroup$
              Says who?$mathstrut$
              $endgroup$
              – Ivan Neretin
              Jan 21 at 8:04






            • 1




              $begingroup$
              @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
              $endgroup$
              – Arthur
              Jan 21 at 8:10












            • $begingroup$
              actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
              $endgroup$
              – choton choton
              Jan 21 at 10:18












            • $begingroup$
              @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
              $endgroup$
              – Arthur
              Jan 21 at 10:23
















            4












            4








            4





            $begingroup$

            Hint: In any group, any element of order $5$ generates a cyclic subgroup of order $5$. Can you think of an invertible linear transformation $T$ of the plane for which $T^5 = I$?






            share|cite|improve this answer









            $endgroup$



            Hint: In any group, any element of order $5$ generates a cyclic subgroup of order $5$. Can you think of an invertible linear transformation $T$ of the plane for which $T^5 = I$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 21 at 7:49









            ArthurArthur

            116k7116198




            116k7116198












            • $begingroup$
              but then entries of the matrix will be complex numbers
              $endgroup$
              – choton choton
              Jan 21 at 7:57






            • 1




              $begingroup$
              Says who?$mathstrut$
              $endgroup$
              – Ivan Neretin
              Jan 21 at 8:04






            • 1




              $begingroup$
              @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
              $endgroup$
              – Arthur
              Jan 21 at 8:10












            • $begingroup$
              actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
              $endgroup$
              – choton choton
              Jan 21 at 10:18












            • $begingroup$
              @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
              $endgroup$
              – Arthur
              Jan 21 at 10:23




















            • $begingroup$
              but then entries of the matrix will be complex numbers
              $endgroup$
              – choton choton
              Jan 21 at 7:57






            • 1




              $begingroup$
              Says who?$mathstrut$
              $endgroup$
              – Ivan Neretin
              Jan 21 at 8:04






            • 1




              $begingroup$
              @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
              $endgroup$
              – Arthur
              Jan 21 at 8:10












            • $begingroup$
              actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
              $endgroup$
              – choton choton
              Jan 21 at 10:18












            • $begingroup$
              @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
              $endgroup$
              – Arthur
              Jan 21 at 10:23


















            $begingroup$
            but then entries of the matrix will be complex numbers
            $endgroup$
            – choton choton
            Jan 21 at 7:57




            $begingroup$
            but then entries of the matrix will be complex numbers
            $endgroup$
            – choton choton
            Jan 21 at 7:57




            1




            1




            $begingroup$
            Says who?$mathstrut$
            $endgroup$
            – Ivan Neretin
            Jan 21 at 8:04




            $begingroup$
            Says who?$mathstrut$
            $endgroup$
            – Ivan Neretin
            Jan 21 at 8:04




            1




            1




            $begingroup$
            @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
            $endgroup$
            – Arthur
            Jan 21 at 8:10






            $begingroup$
            @chotonchoton I never said anything about matrices. I am talking about linear transformations. Granted, the two concepts are quite intimately connected. But one can reason about linear transformations much more intuitively than one can about matrices. And as long as you know that you send the plane to the plane, you know that the corresponding matrix has real entries.
            $endgroup$
            – Arthur
            Jan 21 at 8:10














            $begingroup$
            actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
            $endgroup$
            – choton choton
            Jan 21 at 10:18






            $begingroup$
            actually if there exits a cyclic subgroup of order 5 then there is a invertible martix of order 2 whose order will be 5.But i am not getting any invertible matrix of order 2 whose order is 5
            $endgroup$
            – choton choton
            Jan 21 at 10:18














            $begingroup$
            @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
            $endgroup$
            – Arthur
            Jan 21 at 10:23






            $begingroup$
            @chotonchoton You are using the word order for two different things here. The same matrix can't have order 2 and order 5. But more importantly, I'm trying to tell you to stop thinking about matrices. Think about linear transformations instead. What can you do to the plane so that if you do it five times in a row, you're back where you started?
            $endgroup$
            – Arthur
            Jan 21 at 10:23













            1












            $begingroup$

            Any cyclic group can be regarded as a subgroup of $mathrm{GL}(2, mathbb{R})$. Note that it is always good to regard matrices as linear transformations. For the finite cyclic groups, try to think about a linear transformation that gives you the identity transformation if you apply it for the finite times, say, $n$ times. (Think about the rotation.) For the infinite cyclic group, consider the following matrix:
            $$
            T=begin{pmatrix} 1& 1\0&1end{pmatrix}.
            $$
            How does this matrix act on the real plane $mathbb{R}^{2}$ and why this matrix has infinite order?






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Any cyclic group can be regarded as a subgroup of $mathrm{GL}(2, mathbb{R})$. Note that it is always good to regard matrices as linear transformations. For the finite cyclic groups, try to think about a linear transformation that gives you the identity transformation if you apply it for the finite times, say, $n$ times. (Think about the rotation.) For the infinite cyclic group, consider the following matrix:
              $$
              T=begin{pmatrix} 1& 1\0&1end{pmatrix}.
              $$
              How does this matrix act on the real plane $mathbb{R}^{2}$ and why this matrix has infinite order?






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Any cyclic group can be regarded as a subgroup of $mathrm{GL}(2, mathbb{R})$. Note that it is always good to regard matrices as linear transformations. For the finite cyclic groups, try to think about a linear transformation that gives you the identity transformation if you apply it for the finite times, say, $n$ times. (Think about the rotation.) For the infinite cyclic group, consider the following matrix:
                $$
                T=begin{pmatrix} 1& 1\0&1end{pmatrix}.
                $$
                How does this matrix act on the real plane $mathbb{R}^{2}$ and why this matrix has infinite order?






                share|cite|improve this answer









                $endgroup$



                Any cyclic group can be regarded as a subgroup of $mathrm{GL}(2, mathbb{R})$. Note that it is always good to regard matrices as linear transformations. For the finite cyclic groups, try to think about a linear transformation that gives you the identity transformation if you apply it for the finite times, say, $n$ times. (Think about the rotation.) For the infinite cyclic group, consider the following matrix:
                $$
                T=begin{pmatrix} 1& 1\0&1end{pmatrix}.
                $$
                How does this matrix act on the real plane $mathbb{R}^{2}$ and why this matrix has infinite order?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 8:42









                Seewoo LeeSeewoo Lee

                6,907927




                6,907927






























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