Are the permutations of the Group Caley table rows form the same Group?
$begingroup$
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
asked Dec 22 '18 at 8:30
HHJHHJ
132
$endgroup$
add a comment |
$begingroup$
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
asked Dec 22 '18 at 8:30
HHJHHJ
132
$endgroup$
1
$begingroup$
Yes, this is essentially Cayley's theorem.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 8:34
add a comment |
$begingroup$
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
asked Dec 22 '18 at 8:30
HHJHHJ
132
$endgroup$
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
asked Dec 22 '18 at 8:30
HHJHHJ
132
asked Dec 22 '18 at 8:30
HHJHHJ
132
asked Dec 22 '18 at 8:30
HHJHHJ
132
asked Dec 22 '18 at 8:30
HHJHHJ
132
asked Dec 22 '18 at 8:30
HHJHHJ
132
132
1
$begingroup$
Yes, this is essentially Cayley's theorem.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 8:34
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
$endgroup$
add a comment |
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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.
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
$endgroup$
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.
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
edited Dec 22 '18 at 8:52
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
answered Dec 22 '18 at 8:45
WuestenfuxWuestenfux
4,7331413
4,7331413
add a comment |
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$begingroup$
Yes, this is essentially Cayley's theorem.
$endgroup$
– Lord Shark the Unknown
Dec 22 '18 at 8:34