Is it true that an isomorphism maps elements of the same order to each other?












3












$begingroup$


I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?






abstract-algebra group-theory







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$endgroup$












  • $begingroup$
    One may start with $phi (a^p)=(phi(a))^p$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:44












  • $begingroup$
    do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
    $endgroup$
    – PiccolMan
    Nov 21 '16 at 20:45








  • 1




    $begingroup$
    One has $phi(e_1)=e_2$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:46












  • $begingroup$
    Yes. Order is defined in terms of the language of group theory.
    $endgroup$
    – Alephnull
    Nov 21 '16 at 21:11
















3












$begingroup$


I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    One may start with $phi (a^p)=(phi(a))^p$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:44












  • $begingroup$
    do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
    $endgroup$
    – PiccolMan
    Nov 21 '16 at 20:45








  • 1




    $begingroup$
    One has $phi(e_1)=e_2$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:46












  • $begingroup$
    Yes. Order is defined in terms of the language of group theory.
    $endgroup$
    – Alephnull
    Nov 21 '16 at 21:11














3












3








3


2



$begingroup$


I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?






abstract-algebra group-theory







share|cite|improve this question









$endgroup$




I know that for two groups to be isomorphic each group must contain elements of the same order. So then are elements of the same order mapped to each other? If so, why?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 '16 at 20:42









PiccolManPiccolMan

265110




265110












  • $begingroup$
    One may start with $phi (a^p)=(phi(a))^p$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:44












  • $begingroup$
    do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
    $endgroup$
    – PiccolMan
    Nov 21 '16 at 20:45








  • 1




    $begingroup$
    One has $phi(e_1)=e_2$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:46












  • $begingroup$
    Yes. Order is defined in terms of the language of group theory.
    $endgroup$
    – Alephnull
    Nov 21 '16 at 21:11


















  • $begingroup$
    One may start with $phi (a^p)=(phi(a))^p$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:44












  • $begingroup$
    do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
    $endgroup$
    – PiccolMan
    Nov 21 '16 at 20:45








  • 1




    $begingroup$
    One has $phi(e_1)=e_2$.
    $endgroup$
    – Olivier Oloa
    Nov 21 '16 at 20:46












  • $begingroup$
    Yes. Order is defined in terms of the language of group theory.
    $endgroup$
    – Alephnull
    Nov 21 '16 at 21:11
















$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44






$begingroup$
One may start with $phi (a^p)=(phi(a))^p$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:44














$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45






$begingroup$
do you mean $phi(a^{p}) = (phi(a))^{p}$? If so then this proves it.
$endgroup$
– PiccolMan
Nov 21 '16 at 20:45






1




1




$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46






$begingroup$
One has $phi(e_1)=e_2$.
$endgroup$
– Olivier Oloa
Nov 21 '16 at 20:46














$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11




$begingroup$
Yes. Order is defined in terms of the language of group theory.
$endgroup$
– Alephnull
Nov 21 '16 at 21:11










3 Answers
3






active

oldest

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0












$begingroup$

"Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan



The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Yes,



    because for isomorphism it must hold that $f( u times v) = u cdot v$.



    If order of the element $a$ is $m$ then $a^m = e$



    $f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$



    Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$



    $f$ is an isomorphism, therefore the same argument holds for $f^{-1}$



    $f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$



    $a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$



    Note



    $f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
      $endgroup$
      – PiccolMan
      Nov 21 '16 at 20:50






    • 3




      $begingroup$
      Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
      $endgroup$
      – Stahl
      Nov 21 '16 at 20:54










    • $begingroup$
      @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
      $endgroup$
      – cubeception
      Nov 21 '16 at 21:00






    • 1




      $begingroup$
      @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
      $endgroup$
      – cubeception
      Nov 21 '16 at 21:01



















    0












    $begingroup$

    If a group $G$ contains an element of infinite order then $G$ has infinite order.



    Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$






    share|cite|improve this answer









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      3 Answers
      3






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      3 Answers
      3






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      0












      $begingroup$

      "Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan



      The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        "Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan



        The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          "Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan



          The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.






          share|cite|improve this answer









          $endgroup$



          "Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed." – PiccolMan



          The order of any member of a finite group must have finite order. Let G be a group of order n and a member of G. Look at all a, $a^2$, "$a^3$", $a^4$, ..., $a^{n+1}$. Since there are only n different members of G, at two of those n+1 values must be the same. Suppose $a^i= a^j$ with i>j. Multiplying on both sides by $a^{-j}$, $a^ja^i= a^{i-j}= e$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 '16 at 21:10









          user247327user247327

          10.6k1515




          10.6k1515























              0












              $begingroup$

              Yes,



              because for isomorphism it must hold that $f( u times v) = u cdot v$.



              If order of the element $a$ is $m$ then $a^m = e$



              $f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$



              Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$



              $f$ is an isomorphism, therefore the same argument holds for $f^{-1}$



              $f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$



              $a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$



              Note



              $f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
                $endgroup$
                – PiccolMan
                Nov 21 '16 at 20:50






              • 3




                $begingroup$
                Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
                $endgroup$
                – Stahl
                Nov 21 '16 at 20:54










              • $begingroup$
                @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:00






              • 1




                $begingroup$
                @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:01
















              0












              $begingroup$

              Yes,



              because for isomorphism it must hold that $f( u times v) = u cdot v$.



              If order of the element $a$ is $m$ then $a^m = e$



              $f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$



              Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$



              $f$ is an isomorphism, therefore the same argument holds for $f^{-1}$



              $f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$



              $a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$



              Note



              $f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
                $endgroup$
                – PiccolMan
                Nov 21 '16 at 20:50






              • 3




                $begingroup$
                Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
                $endgroup$
                – Stahl
                Nov 21 '16 at 20:54










              • $begingroup$
                @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:00






              • 1




                $begingroup$
                @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:01














              0












              0








              0





              $begingroup$

              Yes,



              because for isomorphism it must hold that $f( u times v) = u cdot v$.



              If order of the element $a$ is $m$ then $a^m = e$



              $f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$



              Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$



              $f$ is an isomorphism, therefore the same argument holds for $f^{-1}$



              $f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$



              $a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$



              Note



              $f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.






              share|cite|improve this answer











              $endgroup$



              Yes,



              because for isomorphism it must hold that $f( u times v) = u cdot v$.



              If order of the element $a$ is $m$ then $a^m = e$



              $f(a^m)= f(atimes a times dots times a) = f(a) cdot f(a) dots f(a)=f(a)^m=f(e) = e$



              Since $f(a)^m = e$ the order of $f(a)$ is less than or equal to $m$. Let us say that the order of $f(a)$ is $k$, $k le m$



              $f$ is an isomorphism, therefore the same argument holds for $f^{-1}$



              $f^{-1}(f(a)^k)= f^{-1}(f(a) cdot f(a) dots f(a))=f^{-1}(f(a)) cdot f^{-1}(f(a)) dots f^{-1}(f(a)) = a cdot a dots a = a^k = f(e) = e$



              $a^k=e$ therefore $k ge m$. But also $k le m$, so $k=m$



              Note



              $f(e) = e$ because $etimes e= e$ therefore $f(e)cdot f(e)= f(e)$. Therefore $f(e)$ must be the neutral element.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 21 '16 at 21:10

























              answered Nov 21 '16 at 20:49









              cubeceptioncubeception

              1189




              1189












              • $begingroup$
                Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
                $endgroup$
                – PiccolMan
                Nov 21 '16 at 20:50






              • 3




                $begingroup$
                Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
                $endgroup$
                – Stahl
                Nov 21 '16 at 20:54










              • $begingroup$
                @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:00






              • 1




                $begingroup$
                @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:01


















              • $begingroup$
                Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
                $endgroup$
                – PiccolMan
                Nov 21 '16 at 20:50






              • 3




                $begingroup$
                Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
                $endgroup$
                – Stahl
                Nov 21 '16 at 20:54










              • $begingroup$
                @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:00






              • 1




                $begingroup$
                @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
                $endgroup$
                – cubeception
                Nov 21 '16 at 21:01
















              $begingroup$
              Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
              $endgroup$
              – PiccolMan
              Nov 21 '16 at 20:50




              $begingroup$
              Is it possible to have an element of infinite order in a finite group? I think not because then the group would not be closed.
              $endgroup$
              – PiccolMan
              Nov 21 '16 at 20:50




              3




              3




              $begingroup$
              Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
              $endgroup$
              – Stahl
              Nov 21 '16 at 20:54




              $begingroup$
              Your argument shows that the order of $f(a)$ is less than or equal to $m$, not that it equals $m$.
              $endgroup$
              – Stahl
              Nov 21 '16 at 20:54












              $begingroup$
              @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
              $endgroup$
              – cubeception
              Nov 21 '16 at 21:00




              $begingroup$
              @PiccolMan it cannot. If you have a group that has $n$ elements, $a^i$ cannot be different for all $i in {1, dots n}$. Continue from there.
              $endgroup$
              – cubeception
              Nov 21 '16 at 21:00




              1




              1




              $begingroup$
              @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
              $endgroup$
              – cubeception
              Nov 21 '16 at 21:01




              $begingroup$
              @Stahl You are right, but since $f$ is isomorphism one can apply the same argument for $f^{-1}$. Correct?
              $endgroup$
              – cubeception
              Nov 21 '16 at 21:01











              0












              $begingroup$

              If a group $G$ contains an element of infinite order then $G$ has infinite order.



              Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If a group $G$ contains an element of infinite order then $G$ has infinite order.



                Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If a group $G$ contains an element of infinite order then $G$ has infinite order.



                  Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$






                  share|cite|improve this answer









                  $endgroup$



                  If a group $G$ contains an element of infinite order then $G$ has infinite order.



                  Proof: Let $a in G$ with $|a|=infty$. Consider the cyclic group generated by $a$ denoted $langle a rangle$. The order of a cyclic group is the order of a generator of that group, so $|langle a rangle|=|a|=infty$. Thus $langle a rangle$ has infinite order. And $langle a rangle$ is a subgroup of $G$, so we must have $|langle a rangle|<|G|$. $blacksquare$







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                  answered Apr 21 '18 at 17:25









                  M. NestorM. Nestor

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