Show that the factor group $G/N$ is cyclic of order 4.












2












$begingroup$



Let $ G $ be the set of all matrices in $GL_2(mathbb{Z}_5)$ of the form
$ begin{bmatrix}
m & b\
0 & 1
end{bmatrix} $
.
And $N$ is the set of matrices in $G$ of the form $ begin{bmatrix}
1 & c\
0 & 1
end{bmatrix} $
with $cin mathbb{Z}_5$.
Show that the factor group $ G/N $ is cyclic of order 4.




I have already shown that $N$ is a normal subgroup of $G$. But I do not understand what it means for $G/N$ to be cyclic or how to show it. Does that mean that for some $gin G$, $gN$ generates $G/N$? What is the meaning of $gNcdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$



    Let $ G $ be the set of all matrices in $GL_2(mathbb{Z}_5)$ of the form
    $ begin{bmatrix}
    m & b\
    0 & 1
    end{bmatrix} $
    .
    And $N$ is the set of matrices in $G$ of the form $ begin{bmatrix}
    1 & c\
    0 & 1
    end{bmatrix} $
    with $cin mathbb{Z}_5$.
    Show that the factor group $ G/N $ is cyclic of order 4.




    I have already shown that $N$ is a normal subgroup of $G$. But I do not understand what it means for $G/N$ to be cyclic or how to show it. Does that mean that for some $gin G$, $gN$ generates $G/N$? What is the meaning of $gNcdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$



      Let $ G $ be the set of all matrices in $GL_2(mathbb{Z}_5)$ of the form
      $ begin{bmatrix}
      m & b\
      0 & 1
      end{bmatrix} $
      .
      And $N$ is the set of matrices in $G$ of the form $ begin{bmatrix}
      1 & c\
      0 & 1
      end{bmatrix} $
      with $cin mathbb{Z}_5$.
      Show that the factor group $ G/N $ is cyclic of order 4.




      I have already shown that $N$ is a normal subgroup of $G$. But I do not understand what it means for $G/N$ to be cyclic or how to show it. Does that mean that for some $gin G$, $gN$ generates $G/N$? What is the meaning of $gNcdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.










      share|cite|improve this question









      $endgroup$





      Let $ G $ be the set of all matrices in $GL_2(mathbb{Z}_5)$ of the form
      $ begin{bmatrix}
      m & b\
      0 & 1
      end{bmatrix} $
      .
      And $N$ is the set of matrices in $G$ of the form $ begin{bmatrix}
      1 & c\
      0 & 1
      end{bmatrix} $
      with $cin mathbb{Z}_5$.
      Show that the factor group $ G/N $ is cyclic of order 4.




      I have already shown that $N$ is a normal subgroup of $G$. But I do not understand what it means for $G/N$ to be cyclic or how to show it. Does that mean that for some $gin G$, $gN$ generates $G/N$? What is the meaning of $gNcdot gN$? I don't understand what the operation, or its result would be, since $gN$ is a set.







      abstract-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 23:51









      rakaraka

      606




      606






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.



          By definition, we have $gNcdot hN={gncdot hm:n,min N}$

          And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ a homomorphism.



          And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4in N$ ($Leftrightarrow g^4N=eN$)
          but $g^2notin N$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
            $endgroup$
            – raka
            Dec 14 '18 at 0:11












          • $begingroup$
            @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
            $endgroup$
            – Anurag A
            Dec 14 '18 at 0:29





















          0












          $begingroup$

          We know it has order $4$, because $mid Gmid=5^2-5=20$.



          To be cyclic means there's an element of order $4$.



          How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN={gn , | , n in N}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.



            Now to show that $G/N$ is cyclic you need to find an element of the form $gN in G/N$ such that it generates every element of $G/N$. In other words, for all $aN in G/N$, we need to show that there exists a $k in Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.



            In the given case, observe that $ m in {1,2,3,4}$ for the matrices to be invertible and
            $$begin{bmatrix}m&b\0&1end{bmatrix}=
            begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$

            Thus
            $$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
            Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.






            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%2f3038752%2fshow-that-the-factor-group-g-n-is-cyclic-of-order-4%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.



              By definition, we have $gNcdot hN={gncdot hm:n,min N}$

              And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ a homomorphism.



              And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4in N$ ($Leftrightarrow g^4N=eN$)
              but $g^2notin N$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
                $endgroup$
                – raka
                Dec 14 '18 at 0:11












              • $begingroup$
                @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
                $endgroup$
                – Anurag A
                Dec 14 '18 at 0:29


















              0












              $begingroup$

              Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.



              By definition, we have $gNcdot hN={gncdot hm:n,min N}$

              And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ a homomorphism.



              And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4in N$ ($Leftrightarrow g^4N=eN$)
              but $g^2notin N$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
                $endgroup$
                – raka
                Dec 14 '18 at 0:11












              • $begingroup$
                @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
                $endgroup$
                – Anurag A
                Dec 14 '18 at 0:29
















              0












              0








              0





              $begingroup$

              Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.



              By definition, we have $gNcdot hN={gncdot hm:n,min N}$

              And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ a homomorphism.



              And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4in N$ ($Leftrightarrow g^4N=eN$)
              but $g^2notin N$.






              share|cite|improve this answer









              $endgroup$



              Yes, $gN$ is a set, moreover it is a coset of $N$, i.e. an element of $G/N$.



              By definition, we have $gNcdot hN={gncdot hm:n,min N}$

              And, we can see that $gNcdot hN={gh,h^{-1}nhm:n,min N}=ghN$, so the group operation is directly deduced from that of $G$, making the quotient map $Gto G/N$ a homomorphism.



              And yes, $G/N$ being cyclic means that it's a cyclic group, i.e. there's an element $gN$ of it which generates it, which now is equivalent to having order $4$, so that $g^4in N$ ($Leftrightarrow g^4N=eN$)
              but $g^2notin N$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 13 '18 at 23:59









              BerciBerci

              61.2k23674




              61.2k23674












              • $begingroup$
                I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
                $endgroup$
                – raka
                Dec 14 '18 at 0:11












              • $begingroup$
                @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
                $endgroup$
                – Anurag A
                Dec 14 '18 at 0:29




















              • $begingroup$
                I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
                $endgroup$
                – raka
                Dec 14 '18 at 0:11












              • $begingroup$
                @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
                $endgroup$
                – Anurag A
                Dec 14 '18 at 0:29


















              $begingroup$
              I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
              $endgroup$
              – raka
              Dec 14 '18 at 0:11






              $begingroup$
              I feel like it should be obvious but I don't understand why $gNcdot hN = ghN$. What does adding $e=hh^{-1}$ the way you did accomplish? Is it cause we just have $ghn'$ where $n'$ is some arbitrary element of $N$?
              $endgroup$
              – raka
              Dec 14 '18 at 0:11














              $begingroup$
              @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
              $endgroup$
              – Anurag A
              Dec 14 '18 at 0:29






              $begingroup$
              @raka $ghh^{-1}nhm=ghcolor{blue}{h^{-1}nh}color{red}{m}=ghcolor{blue}{underbrace{h^{-1}nh}_{in N}}color{red}{m}$ (using the fact that $N$ is normal in $G$). Now using the fact that $N$is a subgroup, hence closed, we can get $color{blue}{h^{-1}nh}color{red}{m} in N$.
              $endgroup$
              – Anurag A
              Dec 14 '18 at 0:29













              0












              $begingroup$

              We know it has order $4$, because $mid Gmid=5^2-5=20$.



              To be cyclic means there's an element of order $4$.



              How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                We know it has order $4$, because $mid Gmid=5^2-5=20$.



                To be cyclic means there's an element of order $4$.



                How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We know it has order $4$, because $mid Gmid=5^2-5=20$.



                  To be cyclic means there's an element of order $4$.



                  How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?






                  share|cite|improve this answer









                  $endgroup$



                  We know it has order $4$, because $mid Gmid=5^2-5=20$.



                  To be cyclic means there's an element of order $4$.



                  How about $begin{pmatrix}2&1\0&1end{pmatrix}N$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 0:06









                  Chris CusterChris Custer

                  14k3827




                  14k3827























                      0












                      $begingroup$

                      Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN={gn , | , n in N}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.



                      Now to show that $G/N$ is cyclic you need to find an element of the form $gN in G/N$ such that it generates every element of $G/N$. In other words, for all $aN in G/N$, we need to show that there exists a $k in Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.



                      In the given case, observe that $ m in {1,2,3,4}$ for the matrices to be invertible and
                      $$begin{bmatrix}m&b\0&1end{bmatrix}=
                      begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$

                      Thus
                      $$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
                      Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN={gn , | , n in N}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.



                        Now to show that $G/N$ is cyclic you need to find an element of the form $gN in G/N$ such that it generates every element of $G/N$. In other words, for all $aN in G/N$, we need to show that there exists a $k in Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.



                        In the given case, observe that $ m in {1,2,3,4}$ for the matrices to be invertible and
                        $$begin{bmatrix}m&b\0&1end{bmatrix}=
                        begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$

                        Thus
                        $$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
                        Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN={gn , | , n in N}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.



                          Now to show that $G/N$ is cyclic you need to find an element of the form $gN in G/N$ such that it generates every element of $G/N$. In other words, for all $aN in G/N$, we need to show that there exists a $k in Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.



                          In the given case, observe that $ m in {1,2,3,4}$ for the matrices to be invertible and
                          $$begin{bmatrix}m&b\0&1end{bmatrix}=
                          begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$

                          Thus
                          $$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
                          Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.






                          share|cite|improve this answer











                          $endgroup$



                          Starting with a group $G$ and a normal subgroup $N$ you create a partition of the group $G$ based on an equivalence relation. The set of partitions $G/N$ also called the quotient group has elements as cosets $gN={gn , | , n in N}$. This set of partition or quotient group has a group structure (sort of inherited from the group $G$) based on the operation given by $aN cdot bN =(ab)N$. This quotient group can possibly tell us something about the original group $G$.



                          Now to show that $G/N$ is cyclic you need to find an element of the form $gN in G/N$ such that it generates every element of $G/N$. In other words, for all $aN in G/N$, we need to show that there exists a $k in Bbb{Z}$ such that $(gN)^k=aN$, same as $g^kN=aN$.



                          In the given case, observe that $ m in {1,2,3,4}$ for the matrices to be invertible and
                          $$begin{bmatrix}m&b\0&1end{bmatrix}=
                          begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{underbrace{begin{bmatrix}1&bm^{-1}\0&1end{bmatrix}}_{in N}} in begin{bmatrix}m&0\0&1end{bmatrix}color{blue}{N}$$

                          Thus
                          $$G/N=left{begin{bmatrix}m&0\0&1end{bmatrix}N , Big|, m in {1,2,3,4}right}.$$
                          Thus $G/N$ has four elements and now we can see that either $begin{bmatrix}2&0\0&1end{bmatrix}N$ or $begin{bmatrix}3&0\0&1end{bmatrix}N$ can act as generators.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 14 '18 at 0:20

























                          answered Dec 14 '18 at 0:06









                          Anurag AAnurag A

                          26.3k12251




                          26.3k12251






























                              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%2f3038752%2fshow-that-the-factor-group-g-n-is-cyclic-of-order-4%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

                              Plaza Victoria

                              Puebla de Zaragoza

                              Musa