Does the inverse element of a group also have to be under the same set of the group?

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In other words, lets say there is a group $G$. If there is an element $gin G$, by definition there has to be an inverse $g^-1$. Now, my question is, does $g^-1$ have to be an element $in G$? Sorry if this may sound obvious but I want to make sure as the definition doesn't specifically state this.




abstract-algebra group-theory inverse




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  • 5




    Yes, always and forever.
    – Randall
    Aug 27 at 17:07










  • I would also be interested in seeing this definition.
    – Randall
    Aug 27 at 17:07










  • thank you so much
    – mathlover314
    Aug 27 at 17:08






  • 1




    If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
    – Arthur
    Aug 27 at 17:23










  • Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
    – Sambo
    Aug 27 at 19:24














up vote
0
down vote

favorite












In other words, lets say there is a group $G$. If there is an element $gin G$, by definition there has to be an inverse $g^-1$. Now, my question is, does $g^-1$ have to be an element $in G$? Sorry if this may sound obvious but I want to make sure as the definition doesn't specifically state this.




abstract-algebra group-theory inverse




share|cite|improve this question


















  • 5




    Yes, always and forever.
    – Randall
    Aug 27 at 17:07










  • I would also be interested in seeing this definition.
    – Randall
    Aug 27 at 17:07










  • thank you so much
    – mathlover314
    Aug 27 at 17:08






  • 1




    If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
    – Arthur
    Aug 27 at 17:23










  • Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
    – Sambo
    Aug 27 at 19:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











In other words, lets say there is a group $G$. If there is an element $gin G$, by definition there has to be an inverse $g^-1$. Now, my question is, does $g^-1$ have to be an element $in G$? Sorry if this may sound obvious but I want to make sure as the definition doesn't specifically state this.




abstract-algebra group-theory inverse




share|cite|improve this question














In other words, lets say there is a group $G$. If there is an element $gin G$, by definition there has to be an inverse $g^-1$. Now, my question is, does $g^-1$ have to be an element $in G$? Sorry if this may sound obvious but I want to make sure as the definition doesn't specifically state this.





abstract-algebra group-theory inverse





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 27 at 17:31









Sambo

1,3631427




1,3631427










asked Aug 27 at 17:06









mathlover314

41




41







  • 5




    Yes, always and forever.
    – Randall
    Aug 27 at 17:07










  • I would also be interested in seeing this definition.
    – Randall
    Aug 27 at 17:07










  • thank you so much
    – mathlover314
    Aug 27 at 17:08






  • 1




    If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
    – Arthur
    Aug 27 at 17:23










  • Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
    – Sambo
    Aug 27 at 19:24












  • 5




    Yes, always and forever.
    – Randall
    Aug 27 at 17:07










  • I would also be interested in seeing this definition.
    – Randall
    Aug 27 at 17:07










  • thank you so much
    – mathlover314
    Aug 27 at 17:08






  • 1




    If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
    – Arthur
    Aug 27 at 17:23










  • Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
    – Sambo
    Aug 27 at 19:24







5




5




Yes, always and forever.
– Randall
Aug 27 at 17:07




Yes, always and forever.
– Randall
Aug 27 at 17:07












I would also be interested in seeing this definition.
– Randall
Aug 27 at 17:07




I would also be interested in seeing this definition.
– Randall
Aug 27 at 17:07












thank you so much
– mathlover314
Aug 27 at 17:08




thank you so much
– mathlover314
Aug 27 at 17:08




1




1




If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
– Arthur
Aug 27 at 17:23




If your definition doesn't specifically state this, then I would very much like to see exactly what it says.
– Arthur
Aug 27 at 17:23












Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
– Sambo
Aug 27 at 19:24




Arthur and Randall: it's probably just a less formal definition, along the lines of "each element $g$ has an inverse $g^-1$", where the condition that $g^-1 in G$ is implicit.
– Sambo
Aug 27 at 19:24










2 Answers
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Yes, the following is from wikipedia page of Group (mathematics):



For each $a$ in $G$, there exists an element $b$ in $G$, commonly denoted $a^−1$ (or $−a$, if the operation is denoted "$+$"), such that $a cdot b = b cdot a = e$, where $e$ is the identity element.






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    Since the group operation is only required to be on $G$, to make sense of $g cdot g^-1$, we must have $g^-1 in G$.






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      2 Answers
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      2 Answers
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      up vote
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      down vote













      Yes, the following is from wikipedia page of Group (mathematics):



      For each $a$ in $G$, there exists an element $b$ in $G$, commonly denoted $a^−1$ (or $−a$, if the operation is denoted "$+$"), such that $a cdot b = b cdot a = e$, where $e$ is the identity element.






      share|cite|improve this answer
























        up vote
        3
        down vote













        Yes, the following is from wikipedia page of Group (mathematics):



        For each $a$ in $G$, there exists an element $b$ in $G$, commonly denoted $a^−1$ (or $−a$, if the operation is denoted "$+$"), such that $a cdot b = b cdot a = e$, where $e$ is the identity element.






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          Yes, the following is from wikipedia page of Group (mathematics):



          For each $a$ in $G$, there exists an element $b$ in $G$, commonly denoted $a^−1$ (or $−a$, if the operation is denoted "$+$"), such that $a cdot b = b cdot a = e$, where $e$ is the identity element.






          share|cite|improve this answer












          Yes, the following is from wikipedia page of Group (mathematics):



          For each $a$ in $G$, there exists an element $b$ in $G$, commonly denoted $a^−1$ (or $−a$, if the operation is denoted "$+$"), such that $a cdot b = b cdot a = e$, where $e$ is the identity element.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 27 at 17:10









          Siong Thye Goh

          81k1453103




          81k1453103




















              up vote
              1
              down vote













              Since the group operation is only required to be on $G$, to make sense of $g cdot g^-1$, we must have $g^-1 in G$.






              share|cite|improve this answer
























                up vote
                1
                down vote













                Since the group operation is only required to be on $G$, to make sense of $g cdot g^-1$, we must have $g^-1 in G$.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Since the group operation is only required to be on $G$, to make sense of $g cdot g^-1$, we must have $g^-1 in G$.






                  share|cite|improve this answer












                  Since the group operation is only required to be on $G$, to make sense of $g cdot g^-1$, we must have $g^-1 in G$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 27 at 17:14









                  Sambo

                  1,3631427




                  1,3631427



























                       

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