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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081614%2fsubgroup-of-gl2-mathbbr-set-of-all-invertible-matrices-over-mathbbr%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081614%2fsubgroup-of-gl2-mathbbr-set-of-all-invertible-matrices-over-mathbbr%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Mario Kart Wii

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
                    Read more

                    The Binding of Isaac: Rebirth/Afterbirth

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                    Read more

                    What does “Dominus providebit” mean?

                    Image
                    5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
                    Read more