Proof that each element of finite arbitrary group belongs to unique conjugacy class - question on step which takes inverse of a product

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3
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$G$ is an arbitrary finite group. I need to show that each element lies in a unique conjugacy class.



My attempt:



suppose element $g$ lies in conjugacy class $A$ & $B$, but $A neq B$,
then for some $a_x in A, b_x in B, a_x notsim b_x$.



$q, p in G$



$g = p a_x p^-1$



$g=q b_x q^-1$



$a_x=p^-1gp=p^-1q b_x q^-1p$ $(*)$



then define: $m=p^-1q implies m^-1=q^-1p$



Rewrite $(*)$: $a_x=mb_xm^-1$



$m in G$ because $p, q in G implies p^-1, q^-1 in G implies p^-1q, q^-1p in G iff m in G implies a_xsim b_x$
Presupposition was wrong, $implies A = B$.



So there are only disjunct or identical conjugacy classes, so each element belongs to a unique conjugacy class.



Is this proof above correct? If yes, how can I justify the step:



$m=p^-1q implies m^-1=q^-1p$



I know it is true for matrices for example, but is this true for group elements in general?






abstract-algebra group-theory proof-verification finite-groups







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    Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
    – drhab
    1 hour ago














up vote
3
down vote

favorite












$G$ is an arbitrary finite group. I need to show that each element lies in a unique conjugacy class.



My attempt:



suppose element $g$ lies in conjugacy class $A$ & $B$, but $A neq B$,
then for some $a_x in A, b_x in B, a_x notsim b_x$.



$q, p in G$



$g = p a_x p^-1$



$g=q b_x q^-1$



$a_x=p^-1gp=p^-1q b_x q^-1p$ $(*)$



then define: $m=p^-1q implies m^-1=q^-1p$



Rewrite $(*)$: $a_x=mb_xm^-1$



$m in G$ because $p, q in G implies p^-1, q^-1 in G implies p^-1q, q^-1p in G iff m in G implies a_xsim b_x$
Presupposition was wrong, $implies A = B$.



So there are only disjunct or identical conjugacy classes, so each element belongs to a unique conjugacy class.



Is this proof above correct? If yes, how can I justify the step:



$m=p^-1q implies m^-1=q^-1p$



I know it is true for matrices for example, but is this true for group elements in general?






abstract-algebra group-theory proof-verification finite-groups







share|cite|improve this question









New contributor




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















  • 1




    Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
    – drhab
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











$G$ is an arbitrary finite group. I need to show that each element lies in a unique conjugacy class.



My attempt:



suppose element $g$ lies in conjugacy class $A$ & $B$, but $A neq B$,
then for some $a_x in A, b_x in B, a_x notsim b_x$.



$q, p in G$



$g = p a_x p^-1$



$g=q b_x q^-1$



$a_x=p^-1gp=p^-1q b_x q^-1p$ $(*)$



then define: $m=p^-1q implies m^-1=q^-1p$



Rewrite $(*)$: $a_x=mb_xm^-1$



$m in G$ because $p, q in G implies p^-1, q^-1 in G implies p^-1q, q^-1p in G iff m in G implies a_xsim b_x$
Presupposition was wrong, $implies A = B$.



So there are only disjunct or identical conjugacy classes, so each element belongs to a unique conjugacy class.



Is this proof above correct? If yes, how can I justify the step:



$m=p^-1q implies m^-1=q^-1p$



I know it is true for matrices for example, but is this true for group elements in general?






abstract-algebra group-theory proof-verification finite-groups







share|cite|improve this question









New contributor




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











$G$ is an arbitrary finite group. I need to show that each element lies in a unique conjugacy class.



My attempt:



suppose element $g$ lies in conjugacy class $A$ & $B$, but $A neq B$,
then for some $a_x in A, b_x in B, a_x notsim b_x$.



$q, p in G$



$g = p a_x p^-1$



$g=q b_x q^-1$



$a_x=p^-1gp=p^-1q b_x q^-1p$ $(*)$



then define: $m=p^-1q implies m^-1=q^-1p$



Rewrite $(*)$: $a_x=mb_xm^-1$



$m in G$ because $p, q in G implies p^-1, q^-1 in G implies p^-1q, q^-1p in G iff m in G implies a_xsim b_x$
Presupposition was wrong, $implies A = B$.



So there are only disjunct or identical conjugacy classes, so each element belongs to a unique conjugacy class.



Is this proof above correct? If yes, how can I justify the step:



$m=p^-1q implies m^-1=q^-1p$



I know it is true for matrices for example, but is this true for group elements in general?






abstract-algebra group-theory proof-verification finite-groups




abstract-algebra group-theory proof-verification finite-groups






share|cite|improve this question









New contributor




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











share|cite|improve this question









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









share|cite|improve this question




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edited 1 hour ago









José Carlos Santos

123k17101186




123k17101186






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  • 1




    Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
    – drhab
    1 hour ago












  • 1




    Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
    – drhab
    1 hour ago







1




1




Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
– drhab
1 hour ago




Alternative route. The relation can be described by $asim biffexists gin G;bg=ga$. So it is enough to prove that this is an equivalence relation.
– drhab
1 hour ago










2 Answers
2






active

oldest

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up vote
4
down vote



accepted










Yes, it is correct. And, sincebeginalignq^-1poverbracep^-1q^phantomm=m&=q^-1eq\&=q^-1q\&=e,endalignit is indeed true that the inverse of $m$ is $q^-1p$.






share|cite|improve this answer





























    up vote
    2
    down vote













    I'm not sure what the problem is.



    For $x,yin G$, define $xsim y$ if and only if there exists $zin G$ such that
    $$
    y=zxz^-1
    $$
    (that is, $y$ is conjugate to $x$).



    1. The relation $sim$ is reflexive

    2. The relation $sim$ is symmetric

    3. The relation $sim$ is transitive

    The above facts (that you should prove, if you haven't already) say that the relation $sim$ is an equivalence relation.



    If we consider $[x]_sim=yin G:xsim y$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,yin G$, then
    $$
    [x]_sim=[y]_sim qquadtextorqquad [x]_simcap[y]_sim=emptyset
    $$
    that is, two conjugacy classes are either equal or disjoint.



    In particular, no element can belong to two distinct conjugacy classes.






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      Yes, it is correct. And, sincebeginalignq^-1poverbracep^-1q^phantomm=m&=q^-1eq\&=q^-1q\&=e,endalignit is indeed true that the inverse of $m$ is $q^-1p$.






      share|cite|improve this answer


























        up vote
        4
        down vote



        accepted










        Yes, it is correct. And, sincebeginalignq^-1poverbracep^-1q^phantomm=m&=q^-1eq\&=q^-1q\&=e,endalignit is indeed true that the inverse of $m$ is $q^-1p$.






        share|cite|improve this answer
























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Yes, it is correct. And, sincebeginalignq^-1poverbracep^-1q^phantomm=m&=q^-1eq\&=q^-1q\&=e,endalignit is indeed true that the inverse of $m$ is $q^-1p$.






          share|cite|improve this answer














          Yes, it is correct. And, sincebeginalignq^-1poverbracep^-1q^phantomm=m&=q^-1eq\&=q^-1q\&=e,endalignit is indeed true that the inverse of $m$ is $q^-1p$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 20 mins ago

























          answered 1 hour ago









          José Carlos Santos

          123k17101186




          123k17101186




















              up vote
              2
              down vote













              I'm not sure what the problem is.



              For $x,yin G$, define $xsim y$ if and only if there exists $zin G$ such that
              $$
              y=zxz^-1
              $$
              (that is, $y$ is conjugate to $x$).



              1. The relation $sim$ is reflexive

              2. The relation $sim$ is symmetric

              3. The relation $sim$ is transitive

              The above facts (that you should prove, if you haven't already) say that the relation $sim$ is an equivalence relation.



              If we consider $[x]_sim=yin G:xsim y$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,yin G$, then
              $$
              [x]_sim=[y]_sim qquadtextorqquad [x]_simcap[y]_sim=emptyset
              $$
              that is, two conjugacy classes are either equal or disjoint.



              In particular, no element can belong to two distinct conjugacy classes.






              share|cite|improve this answer
























                up vote
                2
                down vote













                I'm not sure what the problem is.



                For $x,yin G$, define $xsim y$ if and only if there exists $zin G$ such that
                $$
                y=zxz^-1
                $$
                (that is, $y$ is conjugate to $x$).



                1. The relation $sim$ is reflexive

                2. The relation $sim$ is symmetric

                3. The relation $sim$ is transitive

                The above facts (that you should prove, if you haven't already) say that the relation $sim$ is an equivalence relation.



                If we consider $[x]_sim=yin G:xsim y$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,yin G$, then
                $$
                [x]_sim=[y]_sim qquadtextorqquad [x]_simcap[y]_sim=emptyset
                $$
                that is, two conjugacy classes are either equal or disjoint.



                In particular, no element can belong to two distinct conjugacy classes.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  I'm not sure what the problem is.



                  For $x,yin G$, define $xsim y$ if and only if there exists $zin G$ such that
                  $$
                  y=zxz^-1
                  $$
                  (that is, $y$ is conjugate to $x$).



                  1. The relation $sim$ is reflexive

                  2. The relation $sim$ is symmetric

                  3. The relation $sim$ is transitive

                  The above facts (that you should prove, if you haven't already) say that the relation $sim$ is an equivalence relation.



                  If we consider $[x]_sim=yin G:xsim y$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,yin G$, then
                  $$
                  [x]_sim=[y]_sim qquadtextorqquad [x]_simcap[y]_sim=emptyset
                  $$
                  that is, two conjugacy classes are either equal or disjoint.



                  In particular, no element can belong to two distinct conjugacy classes.






                  share|cite|improve this answer












                  I'm not sure what the problem is.



                  For $x,yin G$, define $xsim y$ if and only if there exists $zin G$ such that
                  $$
                  y=zxz^-1
                  $$
                  (that is, $y$ is conjugate to $x$).



                  1. The relation $sim$ is reflexive

                  2. The relation $sim$ is symmetric

                  3. The relation $sim$ is transitive

                  The above facts (that you should prove, if you haven't already) say that the relation $sim$ is an equivalence relation.



                  If we consider $[x]_sim=yin G:xsim y$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,yin G$, then
                  $$
                  [x]_sim=[y]_sim qquadtextorqquad [x]_simcap[y]_sim=emptyset
                  $$
                  that is, two conjugacy classes are either equal or disjoint.



                  In particular, no element can belong to two distinct conjugacy classes.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  egreg

                  167k1281189




                  167k1281189




















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