Isomorphic groups with 4 elements [duplicate]












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  • Any group of order four is either cyclic or isomorphic to $V$

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I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?










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marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.























    -1












    $begingroup$



    This question already has an answer here:




    • Any group of order four is either cyclic or isomorphic to $V$

      2 answers




    I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?










    share|cite|improve this question









    $endgroup$



    marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





















      -1












      -1








      -1





      $begingroup$



      This question already has an answer here:




      • Any group of order four is either cyclic or isomorphic to $V$

        2 answers




      I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?










      share|cite|improve this question









      $endgroup$





      This question already has an answer here:




      • Any group of order four is either cyclic or isomorphic to $V$

        2 answers




      I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?





      This question already has an answer here:




      • Any group of order four is either cyclic or isomorphic to $V$

        2 answers








      group-theory






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      asked Jan 13 at 17:26









      GaboruGaboru

      4428




      4428




      marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









      marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























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          If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.






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            0












            $begingroup$

            You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
            1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.






              share|cite|improve this answer









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                2












                $begingroup$

                If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.






                  share|cite|improve this answer









                  $endgroup$



                  If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 17:36









                  user3482749user3482749

                  4,3111019




                  4,3111019























                      0












                      $begingroup$

                      You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
                      1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
                        1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
                          1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)






                          share|cite|improve this answer











                          $endgroup$



                          You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
                          1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 13 at 17:54

























                          answered Jan 13 at 17:43









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