4
$\begingroup$

I would like to permute the elements of the set $A$ using symmetric group. How may I be able to do that? For example, in Sage,

G = SymmetricGroup(3)
for i in G: print [j*i*j^(-1) for j in G]

gives all the conjugacy classes for $S_3$ group. How may I able to do same with SymmetricGroup[3] is Mathematica.

Furthermore, I would like find the elements of $S_3$ that leaves the set $A=\{1, 2\}$ invariant or leaves each element fixed. How may I be able to do that?

$\endgroup$

1 Answer 1

6
$\begingroup$

I find the OP questions somewhat unclear. In any case, the answers can be found in the tutorials "Permutations" and "Permutation Groups".

I would like to permute the elements of the set A using symmetric group.

In[44]:= Permute[{1,2,3},SymmetricGroup[3]]

Out[44]= {{1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1}}

The Mathematica equivalent of the Sage code:

G = SymmetricGroup(3)
for i in G: print [j*i*j^(-1) for j in G]

should be written with other operators/functions, otherwise j*i*j^(-1) will simplify to i. Here I use NonCommutativeMultiply and Inverse:

In[42]:= Permute[NonCommutativeMultiply[j, i, Inverse[j]], SymmetricGroup[3]]

Out[42]= {j ** i ** Inverse[j], j ** Inverse[j] ** i, 
 i ** j ** Inverse[j], i ** Inverse[j] ** j, Inverse[j] ** j ** i, 
 Inverse[j] ** i ** j}

Furthermore, I would like find the elements of $S_3$ that leaves the set $A=\{1,2\}$ invariant [...]

This is unclear to me because:

In[47]:= GroupStabilizer[SymmetricGroup[3], {1, 2}]

Out[47]= PermutationGroup[{}]

hence

In[36]:= Permute[{1,2,3}, GroupStabilizer[SymmetricGroup[3], {1, 2}]]

Out[36]= {{}}

May be the following examples are helpful illustrations.

Fixing the 3d element using $S_3$:

In[50]:= Permute[Range[3], GroupStabilizer[SymmetricGroup[3], {3}]]

Out[50]= {{1, 2, 3}, {2, 1, 3}}

Fixing the elements {1,2} using $S_4$:

In[49]:= Permute[Range[4], GroupStabilizer[SymmetricGroup[4], {1, 2}]]

Out[49]= {{1, 2, 3, 4}, {1, 2, 4, 3}}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.