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In group theory one can calculate some abstract groups through the direct product of two other abstract groups. An example for such a generation is the product $A_5\times Z_2$ with order 120, or $Z_4\times Z_2$ with order 8.

Since the abstract group representation in Mathematica is a permutation group one could have the idea to use the outer product with PermutationProduct on the group elements of two multiplied groups to generate the group product as

GroupOrder[
     PermutationGroup[Flatten[Outer[PermutationProduct,
         GroupElements[grpA],
         GroupElements[grpB]
     ]]]
]

which yields the correct order (120) for grpA = AlternatingGroup[5] and grpB = CyclicGroup[2] but too high an order (24) for grpA = CyclicGroup[4] and grpB = CyclicGroup[2].

One gets the similar result when trying to generate the direct product through using the generators of the two multiplied groups as

GroupOrder[
 PermutationGroup[
  Join[
   GroupGenerators[grpA], 
   GroupGenerators[grpB]
  ]
 ]
]

Any ideas how to solve this issue?

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tl;dr: The cycles of group1 and group2 should not involve the same values. The simplest way to obtain a direct product is to use the function FiniteGroupData with the syntax

FiniteGroupData[ { "DirectProduct", { $group_1$, $group_2$, ...} }, "PermutationGroupRepresentation"]


From the examples

The issue can be found by noticing that $Z_2 \times Z_4$ can be represented as AbelianGroup[{2, 4}], and by comparing the group elements

GroupElements@AbelianGroup[{2, 4}]

(* {Cycles[{}], Cycles[{{3, 4, 5, 6}}], Cycles[{{3, 5}, {4, 6}}], 
    Cycles[{{3, 6, 5, 4}}], Cycles[{{1, 2}}], Cycles[{{1, 2}, {3, 4, 5, 6}}], 
    Cycles[{{1, 2}, {3, 5}, {4, 6}}], Cycles[{{1, 2}, {3, 6, 5, 4}}]} *)

to those obtained from the PermutationProduct of $Z_2$ and $Z_4$

Flatten@ Outer[PermutationProduct,
     GroupElements[CyclicGroup[2]],
     GroupElements[CyclicGroup[4]]]

(* {Cycles[{}], Cycles[{{1, 2, 3, 4}}], Cycles[{{1, 3}, {2, 4}}], 
    Cycles[{{1, 4, 3, 2}}], Cycles[{{1, 2}}], Cycles[{{1, 3, 4}}], 
    Cycles[{{1, 4, 2, 3}}], Cycles[{{2, 4, 3}}]} *)

To get the correct elements, we can replace GroupElements[CyclicGroup[4]] by GroupElements[CyclicGroup[4]] /. Thread[Range[4] -> Range[3, 6]],

cycles = Flatten@Outer[PermutationProduct, 
     GroupElements[CyclicGroup[2]], 
     GroupElements[CyclicGroup[4]] /. Thread[Range[4] -> Range[3, 6]]];

cycles === GroupElements[AbelianGroup[{2, 4}]]

(* True *)

which has the correct group order

GroupOrder[PermutationGroup[cycles]]

(* 8 *)

Generalization

For the direct product of two arbitrary groups, a possible approach could be (see alternative (b) below for a simplest way)

directProduct[group1_, group2_] := With[
     {order1 = GroupOrder[group1], order2 = GroupOrder[group2]},

     PermutationGroup[Flatten@Outer[PermutationProduct,
           GroupElements[group1],
           GroupElements[group2] /. Thread[Range[order2] -> (order1 + Range[order2])]]]
]

For $Z_2 \times Z_4$:

directProduct[CyclicGroup[2], CyclicGroup[4]]
% // GroupOrder

(* PermutationGroup[{Cycles[{}], Cycles[{{3, 4, 5, 6}}], 
  Cycles[{{3, 5}, {4, 6}}], Cycles[{{3, 6, 5, 4}}], Cycles[{{1, 2}}], 
  Cycles[{{1, 2}, {3, 4, 5, 6}}], Cycles[{{1, 2}, {3, 5}, {4, 6}}], 
  Cycles[{{1, 2}, {3, 6, 5, 4}}]}] *)

(* 8 *)

Alternatives

(a) A similar workaround can be applied from the GroupGenerators

directProduct2[group1_, group2_] := PermutationGroup[Join[
     GroupGenerators[group1], 
     GroupGenerators[group2] /. Cycles[l_] :> Cycles[l + PermutationMax[group1]]
]]

For $Z_2 \times Z_4$:

directProduct2[CyclicGroup[2], CyclicGroup[4]]
% // GroupOrder

(* PermutationGroup[{Cycles[{{1, 2}}], Cycles[{{3, 4, 5, 6}}]}] *)
(* 8 *)

In both approaches, one should make sure that the cycles of group1 and those of group2 do not involve the same values.

(b) A simpler way to go, equivalent to alternative (a) above in terms of the cycles generated, is to use the function FiniteGroupData

FiniteGroupData[{"DirectProduct", 
                 {{"CyclicGroup", 2}, {"CyclicGroup", 4}}
                }, "PermutationGroupRepresentation"]

(* PermutationGroup[{Cycles[{{1, 2}}], Cycles[{{3, 4, 5, 6}}]}] *)

FiniteGroupData[{"DirectProduct", 
                 {{"AlternatingGroup", 5}, {"CyclicGroup", 2}}
                }, "PermutationGroupRepresentation"]

(* PermutationGroup[{Cycles[{{6, 7}}], Cycles[{{1, 2, 3}}], Cycles[{{1, 2, 3, 4, 5}}]}] *)
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  • 1
    $\begingroup$ I was not aware of the DirectProduct Property in Finite GroupData. It seems that this is covering even very high order of abstract groups. If I try it with AlternatingGroup of degree 10 and CyclicGroup of degree 2 I get the correct order of 3628800. I'll try out if I find any combination of abstract groups which is not covered with this, but it seems that FiniteGroupData does the trick, thanks! $\endgroup$ – Rainer Mar 13 '16 at 9:16
  • $\begingroup$ one additional item @Xavier. Where did you find the FiniteGroupData command syntax on the DirectProduct? I checked with the MMA documentation and did not find it there.... $\endgroup$ – Rainer Mar 13 '16 at 9:36
  • $\begingroup$ @Rainer You can find it in the "Details" section of FiniteGroupData, second Table. Three special group specifications are shown: "AbelianGroup", "DirectProduct" and "SemidirectProduct". $\endgroup$ – user31159 Mar 13 '16 at 13:27

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