Do isomorphic groups share a binary operation?












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Suppose you have two isomorphic groups. Do they have to share a binary operation?






group-theory group-homomorphism binary-operations







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  • $begingroup$
    If you mean, does it have to be the same operation in both groups: no.
    $endgroup$
    – David
    3 hours ago










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    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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    – Shaun
    2 hours ago










  • $begingroup$
    Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
    $endgroup$
    – Ethan Bolker
    15 mins ago
















2












$begingroup$


Suppose you have two isomorphic groups. Do they have to share a binary operation?






group-theory group-homomorphism binary-operations







share|cite|improve this question









New contributor




NickBlotted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    If you mean, does it have to be the same operation in both groups: no.
    $endgroup$
    – David
    3 hours ago










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Shaun
    2 hours ago










  • $begingroup$
    Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
    $endgroup$
    – Ethan Bolker
    15 mins ago














2












2








2


1



$begingroup$


Suppose you have two isomorphic groups. Do they have to share a binary operation?






group-theory group-homomorphism binary-operations







share|cite|improve this question









New contributor




NickBlotted is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Suppose you have two isomorphic groups. Do they have to share a binary operation?






group-theory group-homomorphism binary-operations




group-theory group-homomorphism binary-operations






share|cite|improve this question









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edited 2 hours ago









Shaun

9,000113682




9,000113682






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asked 3 hours ago









NickBlottedNickBlotted

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  • $begingroup$
    If you mean, does it have to be the same operation in both groups: no.
    $endgroup$
    – David
    3 hours ago










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Shaun
    2 hours ago










  • $begingroup$
    Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
    $endgroup$
    – Ethan Bolker
    15 mins ago


















  • $begingroup$
    If you mean, does it have to be the same operation in both groups: no.
    $endgroup$
    – David
    3 hours ago










  • $begingroup$
    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
    $endgroup$
    – Shaun
    2 hours ago










  • $begingroup$
    Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
    $endgroup$
    – Ethan Bolker
    15 mins ago
















$begingroup$
If you mean, does it have to be the same operation in both groups: no.
$endgroup$
– David
3 hours ago




$begingroup$
If you mean, does it have to be the same operation in both groups: no.
$endgroup$
– David
3 hours ago












$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
2 hours ago




$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Shaun
2 hours ago












$begingroup$
Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
$endgroup$
– Ethan Bolker
15 mins ago




$begingroup$
Short answer" "no". Possibly helpful: math.stackexchange.com/questions/2039702/…
$endgroup$
– Ethan Bolker
15 mins ago










4 Answers
4






active

oldest

votes


















3












$begingroup$

$G_1 = {0}$ and $G_2 = {1}$ are groups (each with multiplication as the binary operation).



They are isomorphic, as you can show.



The binary operation of $G_1$ has as its domain ${(0,0)}$, while the binary operation of $G_2$ has as its domain ${(1,1)}$, so their binary operations are not equal.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Not necessarily, no.



    The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.



    The underlying sets can be completely different too. A bijection between sets is not necessarily an equality; think: permutations.






    share|cite|improve this answer











    $endgroup$





















      3












      $begingroup$

      Consider this:



      $log: (mathbb R^+,times) to (mathbb R,+)$ is a group isomorphism.






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        To add yet another example to those above, note that if $mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $sigma longmapsto P_sigma$.






        share|cite|improve this answer











        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $G_1 = {0}$ and $G_2 = {1}$ are groups (each with multiplication as the binary operation).



          They are isomorphic, as you can show.



          The binary operation of $G_1$ has as its domain ${(0,0)}$, while the binary operation of $G_2$ has as its domain ${(1,1)}$, so their binary operations are not equal.






          share|cite|improve this answer









          $endgroup$


















            3












            $begingroup$

            $G_1 = {0}$ and $G_2 = {1}$ are groups (each with multiplication as the binary operation).



            They are isomorphic, as you can show.



            The binary operation of $G_1$ has as its domain ${(0,0)}$, while the binary operation of $G_2$ has as its domain ${(1,1)}$, so their binary operations are not equal.






            share|cite|improve this answer









            $endgroup$
















              3












              3








              3





              $begingroup$

              $G_1 = {0}$ and $G_2 = {1}$ are groups (each with multiplication as the binary operation).



              They are isomorphic, as you can show.



              The binary operation of $G_1$ has as its domain ${(0,0)}$, while the binary operation of $G_2$ has as its domain ${(1,1)}$, so their binary operations are not equal.






              share|cite|improve this answer









              $endgroup$



              $G_1 = {0}$ and $G_2 = {1}$ are groups (each with multiplication as the binary operation).



              They are isomorphic, as you can show.



              The binary operation of $G_1$ has as its domain ${(0,0)}$, while the binary operation of $G_2$ has as its domain ${(1,1)}$, so their binary operations are not equal.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              MetricMetric

              1,20139




              1,20139























                  3












                  $begingroup$

                  Not necessarily, no.



                  The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.



                  The underlying sets can be completely different too. A bijection between sets is not necessarily an equality; think: permutations.






                  share|cite|improve this answer











                  $endgroup$


















                    3












                    $begingroup$

                    Not necessarily, no.



                    The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.



                    The underlying sets can be completely different too. A bijection between sets is not necessarily an equality; think: permutations.






                    share|cite|improve this answer











                    $endgroup$
















                      3












                      3








                      3





                      $begingroup$

                      Not necessarily, no.



                      The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.



                      The underlying sets can be completely different too. A bijection between sets is not necessarily an equality; think: permutations.






                      share|cite|improve this answer











                      $endgroup$



                      Not necessarily, no.



                      The two binary operations behave the same on the underlying sets with respect to the axioms of group theory, yeah, by the fact that the two groups are isomorphic as groups, but it might so happen that, say, one binary operation is the restriction of some bigger relation whereas the other is not.



                      The underlying sets can be completely different too. A bijection between sets is not necessarily an equality; think: permutations.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 hours ago

























                      answered 2 hours ago









                      ShaunShaun

                      9,000113682




                      9,000113682























                          3












                          $begingroup$

                          Consider this:



                          $log: (mathbb R^+,times) to (mathbb R,+)$ is a group isomorphism.






                          share|cite|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            Consider this:



                            $log: (mathbb R^+,times) to (mathbb R,+)$ is a group isomorphism.






                            share|cite|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Consider this:



                              $log: (mathbb R^+,times) to (mathbb R,+)$ is a group isomorphism.






                              share|cite|improve this answer









                              $endgroup$



                              Consider this:



                              $log: (mathbb R^+,times) to (mathbb R,+)$ is a group isomorphism.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              lhflhf

                              164k10170395




                              164k10170395























                                  0












                                  $begingroup$

                                  To add yet another example to those above, note that if $mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $sigma longmapsto P_sigma$.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    0












                                    $begingroup$

                                    To add yet another example to those above, note that if $mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $sigma longmapsto P_sigma$.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      To add yet another example to those above, note that if $mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $sigma longmapsto P_sigma$.






                                      share|cite|improve this answer











                                      $endgroup$



                                      To add yet another example to those above, note that if $mathcal{P}_n$ denotes the group of $n$-by-$n$ permutation matrices (binary operation is matrix multiplication) and $S_n$ denotes the symmetric group (binary operation is function-composition), then these groups are isomorphic via the isomorphism $sigma longmapsto P_sigma$.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 16 mins ago

























                                      answered 22 mins ago









                                      Pietro PaparellaPietro Paparella

                                      1,372615




                                      1,372615






















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