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I would like to define a group of order 12 in Mathematica by giving a list of 12 elements in cycle notation,

$$ S = \{(e), (123), (132), (12)(45), (13)(45), (23)(45), (13), (23), (12), (45), (123)(45), (132)(45) \}$$

Is such a thing possible? If so, how? If I even so much as try to define a group using Cycles I get an error because as written above, the cycles do not have distinct elements.

I would like to know if there's a way to then do stuff like finding the character table of this group, using Mathematica.


For example, the following line

PermutationGroup[Cycles[{{1, 2, 3}, {1, 3, 2}, {1, 3}, {2, 3}, {1, 2}}]]

produces an error

Cycles::reppoint: Cycles[{{1,2,3},{1,3,2},{1,3},{2,3},{1,2}}] contains repeated integers.

By the way, this isn't even a specification of the full group. How does one specify products of cycles, like $(12)(45)$ or $(123)(45)$?

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    $\begingroup$ Could you include the code where you used PermutationGroup[] and Cycles[] to specify the group? It seems to me that you might have just made a syntax error somewhere. $\endgroup$ – J. M. is away Oct 15 '18 at 21:35
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    $\begingroup$ You should be feeding a list of Cycles[] to PermutationGroup[], e.g. PermutationGroup[{Cycles[{}], Cycles[{{1, 3, 2}}], Cycles[{{1, 2}, {4, 5}}]}] $\endgroup$ – J. M. is away Oct 15 '18 at 21:41
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cycles = {{{}}, {{1, 2, 3}}, {{1, 3, 2}}, {{1, 2}, {4, 5}}, {{1, 3}, {4, 5}}, 
 {{2, 3}, {4, 5}}, {{1, 3}}, {{2, 3}}, {{1, 2}}, {{4, 5}}, 
 {{1, 2, 3}, {4, 5}}, {{1, 3, 2}, {4, 5}}};
pg = PermutationGroup[Cycles /@ cycles];
GroupOrder[pg]

12

GroupMultiplicationTable[pg] // MatrixForm // TeXForm

$\left( \begin{array}{cccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 2 & 1 & 4 & 3 & 6 & 5 & 8 & 7 & 10 & 9 & 12 & 11 \\ 3 & 4 & 1 & 2 & 7 & 8 & 5 & 6 & 11 & 12 & 9 & 10 \\ 4 & 3 & 2 & 1 & 8 & 7 & 6 & 5 & 12 & 11 & 10 & 9 \\ 5 & 6 & 9 & 10 & 1 & 2 & 11 & 12 & 3 & 4 & 7 & 8 \\ 6 & 5 & 10 & 9 & 2 & 1 & 12 & 11 & 4 & 3 & 8 & 7 \\ 7 & 8 & 11 & 12 & 3 & 4 & 9 & 10 & 1 & 2 & 5 & 6 \\ 8 & 7 & 12 & 11 & 4 & 3 & 10 & 9 & 2 & 1 & 6 & 5 \\ 9 & 10 & 5 & 6 & 11 & 12 & 1 & 2 & 7 & 8 & 3 & 4 \\ 10 & 9 & 6 & 5 & 12 & 11 & 2 & 1 & 8 & 7 & 4 & 3 \\ 11 & 12 & 7 & 8 & 9 & 10 & 3 & 4 & 5 & 6 & 1 & 2 \\ 12 & 11 & 8 & 7 & 10 & 9 & 4 & 3 & 6 & 5 & 2 & 1 \\ \end{array} \right)$

PermutationCycles /@ GroupMultiplicationTable[pg]

{Cycles[{}], Cycles[{{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}, {11, 12}}],
Cycles[{{1, 3}, {2, 4}, {5, 7}, {6, 8}, {9, 11}, {10, 12}}],
Cycles[{{1, 4}, {2, 3}, {5, 8}, {6, 7}, {9, 12}, {10, 11}}],
Cycles[{{1, 5}, {2, 6}, {3, 9}, {4, 10}, {7, 11}, {8, 12}}],
Cycles[{{1, 6}, {2, 5}, {3, 10}, {4, 9}, {7, 12}, {8, 11}}],
Cycles[{{1, 7, 9}, {2, 8, 10}, {3, 11, 5}, {4, 12, 6}}],
Cycles[{{1, 8, 9, 2, 7, 10}, {3, 12, 5, 4, 11, 6}}],
Cycles[{{1, 9, 7}, {2, 10, 8}, {3, 5, 11}, {4, 6, 12}}],
Cycles[{{1, 10, 7, 2, 9, 8}, {3, 6, 11, 4, 5, 12}}],
Cycles[{{1, 11}, {2, 12}, {3, 7}, {4, 8}, {5, 9}, {6, 10}}],
Cycles[{{1, 12}, {2, 11}, {3, 8}, {4, 7}, {5, 10}, {6, 9}}]}

 PermutationProduct[Cycles[{{1, 2}, {4, 5}}],  Cycles[{{1, 2, 3}, {4, 5}}]]

Cycles[{{1, 3}}]

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  • $\begingroup$ Don't you want it to be cycles = {{{}}, {{1, 2, 3}}, {{1, 3, 2}}, {{1, 2}, {4, 5}}, {{1, 3}, {4, 5}}, {{2, 3}, {4, 5}}, {{1, 3}}, {{2, 3}}, {{1, 2}}, {{4, 5}}, {{1, 2, 3}, {4, 5}}, {{1, 3, 2}, {4, 5}}}; ? $\endgroup$ – leastaction Oct 15 '18 at 21:54
  • $\begingroup$ @leastaction, yes; missed few elements in your list. Corrected now. $\endgroup$ – kglr Oct 15 '18 at 21:57
  • $\begingroup$ You might also be interested in GroupMultiplicationTable[] @least. $\endgroup$ – J. M. is away Oct 15 '18 at 22:05
  • $\begingroup$ Thanks, yes. I am -- but can we re-express the elements in cycle form? $\endgroup$ – leastaction Oct 15 '18 at 22:05

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