Let S= {"A","B","C","D"} and S4= SymmetricGroup(4). I want to create a table of the action S4 x S -> S which standardly permutes the letters in the set. The table should look like:

Permutation.   A.  B.  C.  D
()             A.  B.  C.  D
(1 2 )         B.  A.  C.  D.

How can this be done with Sage? ( The Set and Group in the question are just examples, I want to be able to create a table for any action.

  • 1
    $\begingroup$ If you don't get answers here, you could try asking here or here. $\endgroup$ – Fredrik Meyer Jan 2 '15 at 13:28
  • $\begingroup$ Thanks, I will definitely try that, I used to code a lot in Mathematica for which I have had so much excellent support here at Mathematica SE. I hoped to get some Sage support on SE but I think you are right: this might not be the place, despite the tag. I can do most of the stuff directly in GAP but I want to learn it the Sage way. Only then I can decide if Sage is what I need. $\endgroup$ – nilo de roock Jan 2 '15 at 13:44
  • 1
    $\begingroup$ Here is the sage-support thread groups.google.com/forum/#!topic/sage-support/20gEbPVFd_Y $\endgroup$ – kcrisman Jan 2 '15 at 15:52

Given a group $G$, an ordered set $X$ and an action $\varphi: G \times X \to X$, based on your example you want a table with rows $g, \operatorname{im} \varphi(g, \cdot)$, where the images of the elements of $X$ are computed in the given order.

We can arrange this in SAGE as follows (using your example):

G = SymmetricGroup(4)
X = ['A','B','C','D']
act = lambda g, x: X[g(X.index(x) + 1) - 1]
im = lambda g: [act(g,x) for x in X]

table([(g, im(g)) for g in G])


()           ['A', 'B', 'C', 'D']
(3,4)        ['A', 'B', 'D', 'C']
(2,3)        ['A', 'C', 'B', 'D']
(2,3,4)      ['A', 'C', 'D', 'B']
(2,4,3)      ['A', 'D', 'B', 'C']
(2,4)        ['A', 'D', 'C', 'B']
(1,2)        ['B', 'A', 'C', 'D']
(1,2)(3,4)   ['B', 'A', 'D', 'C']
(1,2,3)      ['B', 'C', 'A', 'D']
(1,2,3,4)    ['B', 'C', 'D', 'A']
(1,2,4,3)    ['B', 'D', 'A', 'C']
(1,2,4)      ['B', 'D', 'C', 'A']
(1,3,2)      ['C', 'A', 'B', 'D']
(1,3,4,2)    ['C', 'A', 'D', 'B']
(1,3)        ['C', 'B', 'A', 'D']
(1,3,4)      ['C', 'B', 'D', 'A']
(1,3)(2,4)   ['C', 'D', 'A', 'B']
(1,3,2,4)    ['C', 'D', 'B', 'A']
(1,4,3,2)    ['D', 'A', 'B', 'C']
(1,4,2)      ['D', 'A', 'C', 'B']
(1,4,3)      ['D', 'B', 'A', 'C']
(1,4)        ['D', 'B', 'C', 'A']
(1,4,2,3)    ['D', 'C', 'A', 'B']
(1,4)(2,3)   ['D', 'C', 'B', 'A']

The row with the identity element conveniently serves as a heading.

Clearly hardest part of this is to define the action.

| cite | improve this answer | |
  • 1
    $\begingroup$ ( Dank je wel Ricardo) Thanks, it is the answer I was looking for. $\endgroup$ – nilo de roock Jan 2 '15 at 17:41
  • $\begingroup$ Damn!! I was looking for it everywhere. $\endgroup$ – Santosh Linkha Sep 8 '16 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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