There are only two groups of order six, up to isomorphism: $mathbb Z_6$ and $S_3$. [duplicate]












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  • There are 2 groups of order 6 (up to isomorphism) [duplicate]

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Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.




I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.










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marked as duplicate by Chinnapparaj R, jgon, DRF, A. Pongrácz, amWhy abstract-algebra
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Dec 4 '18 at 18:23


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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    This question already has an answer here:




    • There are 2 groups of order 6 (up to isomorphism) [duplicate]

      2 answers





    Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.




    I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.










    share|cite|improve this question











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    marked as duplicate by Chinnapparaj R, jgon, DRF, A. Pongrácz, amWhy abstract-algebra
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    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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      This question already has an answer here:




      • There are 2 groups of order 6 (up to isomorphism) [duplicate]

        2 answers





      Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.




      I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.










      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • There are 2 groups of order 6 (up to isomorphism) [duplicate]

        2 answers





      Let $G$ be group with order $6$. Prove that either $G$ and $Bbb Z_{6}$ are isomorphic binary structure or $G$ and $S_{3}$ are isomorphic binary structure.




      I know that for isomorphic binary structure, we define a function between groups and we should check homomorphism property and bijection. But I can not define a function between them. Please help me, if you have any good idea.





      This question already has an answer here:




      • There are 2 groups of order 6 (up to isomorphism) [duplicate]

        2 answers








      abstract-algebra group-theory finite-groups group-isomorphism






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      edited Dec 4 '18 at 22:50









      amWhy

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      asked Dec 4 '18 at 9:32









      mathsstudentmathsstudent

      383




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      marked as duplicate by Chinnapparaj R, jgon, DRF, A. Pongrácz, amWhy abstract-algebra
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      Dec 4 '18 at 18:23


      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 Chinnapparaj R, jgon, DRF, A. Pongrácz, amWhy abstract-algebra
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      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 Answer
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          0












          $begingroup$

          Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:48










          • $begingroup$
            @amWhy This time, I did search for a duplicate.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:50










          • $begingroup$
            Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:51












          • $begingroup$
            @amWhy I had never thought of that. That's a good idea.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:52


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:48










          • $begingroup$
            @amWhy This time, I did search for a duplicate.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:50










          • $begingroup$
            Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:51












          • $begingroup$
            @amWhy I had never thought of that. That's a good idea.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:52
















          0












          $begingroup$

          Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:48










          • $begingroup$
            @amWhy This time, I did search for a duplicate.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:50










          • $begingroup$
            Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:51












          • $begingroup$
            @amWhy I had never thought of that. That's a good idea.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:52














          0












          0








          0





          $begingroup$

          Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?






          share|cite|improve this answer









          $endgroup$



          Hint: Either your group has an element of order $6$ or not. If it has, then it is isomorphic to $mathbb{Z}_6$. Otherwise, which are the possible orders of elemets of $G$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 9:40









          José Carlos SantosJosé Carlos Santos

          155k22124227




          155k22124227








          • 1




            $begingroup$
            Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:48










          • $begingroup$
            @amWhy This time, I did search for a duplicate.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:50










          • $begingroup$
            Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:51












          • $begingroup$
            @amWhy I had never thought of that. That's a good idea.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:52














          • 1




            $begingroup$
            Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:48










          • $begingroup$
            @amWhy This time, I did search for a duplicate.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:50










          • $begingroup$
            Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
            $endgroup$
            – amWhy
            Dec 4 '18 at 22:51












          • $begingroup$
            @amWhy I had never thought of that. That's a good idea.
            $endgroup$
            – José Carlos Santos
            Dec 4 '18 at 22:52








          1




          1




          $begingroup$
          Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
          $endgroup$
          – amWhy
          Dec 4 '18 at 22:48




          $begingroup$
          Please start at least cursory efforts to look for duplicates of questions before you answer them. This is a rather standard question which you'd expect to have been asked on this site before, multiple times.
          $endgroup$
          – amWhy
          Dec 4 '18 at 22:48












          $begingroup$
          @amWhy This time, I did search for a duplicate.
          $endgroup$
          – José Carlos Santos
          Dec 4 '18 at 22:50




          $begingroup$
          @amWhy This time, I did search for a duplicate.
          $endgroup$
          – José Carlos Santos
          Dec 4 '18 at 22:50












          $begingroup$
          Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
          $endgroup$
          – amWhy
          Dec 4 '18 at 22:51






          $begingroup$
          Okay, I'm fine with that. I changed the title of the question so that it, like its duplicate, is more readily accessible in the future for similar questions.
          $endgroup$
          – amWhy
          Dec 4 '18 at 22:51














          $begingroup$
          @amWhy I had never thought of that. That's a good idea.
          $endgroup$
          – José Carlos Santos
          Dec 4 '18 at 22:52




          $begingroup$
          @amWhy I had never thought of that. That's a good idea.
          $endgroup$
          – José Carlos Santos
          Dec 4 '18 at 22:52



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