Are the permutations of the Group Caley table rows form the same Group?












0












$begingroup$


For Example :-
Z/Z4



0 1 2 3 -> ()

1 2 3 0 -> (0, 1, 2, 3)

2 3 0 1 -> (0, 2)(1, 3)

3 0 1 2 -> (0, 3, 2, 1)



The above Permutations under permutation combination operation also form the same group.



Example of Nonabelian Group:-
Group D_3



0 1 2 3 4 5 -> ()

1 2 0 4 5 3 -> (0 ,1 ,2)(3 ,4 ,5)

2 0 1 5 3 4 -> (0 ,2 ,1)(3 ,5 ,4)

3 5 4 0 2 1 -> (0 ,3)(1 ,5)(2 ,4)

4 3 5 1 0 2 -> (0 ,4)(1 ,3)(2 ,5)

5 4 3 2 1 0 -> (0 ,5)(1 ,4)(2 ,3)



The above Permutations under permutation combination operation also form the same group.



I haven't found counter example yet hence I'm wondering if Is this is true for all Groups? and can this relation be proven?






abstract-algebra group-theory finite-groups permutations







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$endgroup$








  • 1




    $begingroup$
    Yes, this is essentially Cayley's theorem.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 8:34
















0












$begingroup$


For Example :-
Z/Z4



0 1 2 3 -> ()

1 2 3 0 -> (0, 1, 2, 3)

2 3 0 1 -> (0, 2)(1, 3)

3 0 1 2 -> (0, 3, 2, 1)



The above Permutations under permutation combination operation also form the same group.



Example of Nonabelian Group:-
Group D_3



0 1 2 3 4 5 -> ()

1 2 0 4 5 3 -> (0 ,1 ,2)(3 ,4 ,5)

2 0 1 5 3 4 -> (0 ,2 ,1)(3 ,5 ,4)

3 5 4 0 2 1 -> (0 ,3)(1 ,5)(2 ,4)

4 3 5 1 0 2 -> (0 ,4)(1 ,3)(2 ,5)

5 4 3 2 1 0 -> (0 ,5)(1 ,4)(2 ,3)



The above Permutations under permutation combination operation also form the same group.



I haven't found counter example yet hence I'm wondering if Is this is true for all Groups? and can this relation be proven?






abstract-algebra group-theory finite-groups permutations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, this is essentially Cayley's theorem.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 8:34














0












0








0





$begingroup$


For Example :-
Z/Z4



0 1 2 3 -> ()

1 2 3 0 -> (0, 1, 2, 3)

2 3 0 1 -> (0, 2)(1, 3)

3 0 1 2 -> (0, 3, 2, 1)



The above Permutations under permutation combination operation also form the same group.



Example of Nonabelian Group:-
Group D_3



0 1 2 3 4 5 -> ()

1 2 0 4 5 3 -> (0 ,1 ,2)(3 ,4 ,5)

2 0 1 5 3 4 -> (0 ,2 ,1)(3 ,5 ,4)

3 5 4 0 2 1 -> (0 ,3)(1 ,5)(2 ,4)

4 3 5 1 0 2 -> (0 ,4)(1 ,3)(2 ,5)

5 4 3 2 1 0 -> (0 ,5)(1 ,4)(2 ,3)



The above Permutations under permutation combination operation also form the same group.



I haven't found counter example yet hence I'm wondering if Is this is true for all Groups? and can this relation be proven?






abstract-algebra group-theory finite-groups permutations







share|cite|improve this question









$endgroup$




For Example :-
Z/Z4



0 1 2 3 -> ()

1 2 3 0 -> (0, 1, 2, 3)

2 3 0 1 -> (0, 2)(1, 3)

3 0 1 2 -> (0, 3, 2, 1)



The above Permutations under permutation combination operation also form the same group.



Example of Nonabelian Group:-
Group D_3



0 1 2 3 4 5 -> ()

1 2 0 4 5 3 -> (0 ,1 ,2)(3 ,4 ,5)

2 0 1 5 3 4 -> (0 ,2 ,1)(3 ,5 ,4)

3 5 4 0 2 1 -> (0 ,3)(1 ,5)(2 ,4)

4 3 5 1 0 2 -> (0 ,4)(1 ,3)(2 ,5)

5 4 3 2 1 0 -> (0 ,5)(1 ,4)(2 ,3)



The above Permutations under permutation combination operation also form the same group.



I haven't found counter example yet hence I'm wondering if Is this is true for all Groups? and can this relation be proven?






abstract-algebra group-theory finite-groups permutations




abstract-algebra group-theory finite-groups permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Dec 22 '18 at 8:30









HHJHHJ

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132








  • 1




    $begingroup$
    Yes, this is essentially Cayley's theorem.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 8:34














  • 1




    $begingroup$
    Yes, this is essentially Cayley's theorem.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 8:34








1




1




$begingroup$
Yes, this is essentially Cayley's theorem.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 8:34




$begingroup$
Yes, this is essentially Cayley's theorem.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 8:34










1 Answer
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$begingroup$

Well, given a group $G={g_1,ldots,g_n}$, consider the left multiplications $l_{g_i}:Grightarrow G: gmapsto g_ig$. The left multiplications correspond to the rows in the group table.



The set of left multiplications $L={l_{g_1},ldots,l_{g_n}}$ forms a subgroup of the symmetric group $S_G={fmid f:Grightarrow Gmbox{ bij.}}$, since $l_{g}l_h=l_{gh}$ and $l_g^{-1}=l_{g^{-1}}$.



Thus the left multiplications are permutations of the set $G$ and so have a representation in cycle form.



Have a look into Cayley's theorem.






share|cite|improve this answer











$endgroup$













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    $begingroup$

    Well, given a group $G={g_1,ldots,g_n}$, consider the left multiplications $l_{g_i}:Grightarrow G: gmapsto g_ig$. The left multiplications correspond to the rows in the group table.



    The set of left multiplications $L={l_{g_1},ldots,l_{g_n}}$ forms a subgroup of the symmetric group $S_G={fmid f:Grightarrow Gmbox{ bij.}}$, since $l_{g}l_h=l_{gh}$ and $l_g^{-1}=l_{g^{-1}}$.



    Thus the left multiplications are permutations of the set $G$ and so have a representation in cycle form.



    Have a look into Cayley's theorem.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Well, given a group $G={g_1,ldots,g_n}$, consider the left multiplications $l_{g_i}:Grightarrow G: gmapsto g_ig$. The left multiplications correspond to the rows in the group table.



      The set of left multiplications $L={l_{g_1},ldots,l_{g_n}}$ forms a subgroup of the symmetric group $S_G={fmid f:Grightarrow Gmbox{ bij.}}$, since $l_{g}l_h=l_{gh}$ and $l_g^{-1}=l_{g^{-1}}$.



      Thus the left multiplications are permutations of the set $G$ and so have a representation in cycle form.



      Have a look into Cayley's theorem.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Well, given a group $G={g_1,ldots,g_n}$, consider the left multiplications $l_{g_i}:Grightarrow G: gmapsto g_ig$. The left multiplications correspond to the rows in the group table.



        The set of left multiplications $L={l_{g_1},ldots,l_{g_n}}$ forms a subgroup of the symmetric group $S_G={fmid f:Grightarrow Gmbox{ bij.}}$, since $l_{g}l_h=l_{gh}$ and $l_g^{-1}=l_{g^{-1}}$.



        Thus the left multiplications are permutations of the set $G$ and so have a representation in cycle form.



        Have a look into Cayley's theorem.






        share|cite|improve this answer











        $endgroup$



        Well, given a group $G={g_1,ldots,g_n}$, consider the left multiplications $l_{g_i}:Grightarrow G: gmapsto g_ig$. The left multiplications correspond to the rows in the group table.



        The set of left multiplications $L={l_{g_1},ldots,l_{g_n}}$ forms a subgroup of the symmetric group $S_G={fmid f:Grightarrow Gmbox{ bij.}}$, since $l_{g}l_h=l_{gh}$ and $l_g^{-1}=l_{g^{-1}}$.



        Thus the left multiplications are permutations of the set $G$ and so have a representation in cycle form.



        Have a look into Cayley's theorem.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 22 '18 at 8:52

























        answered Dec 22 '18 at 8:45









        WuestenfuxWuestenfux

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        4,7331413






























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