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I am interested in obtaining a permutation representation of the Pauli Group $G_1 = \langle X, Y, Z \rangle$. I think this would be easy enough as a "Regular representation" but then I learn that the Pauli group is the central product of $C_4$ and $D_4$. The definition of "central product" seems to reside in the darker recesses of group theory and I don't understand the terse notation I have found for it.

If someone could provide the Mathematica code for the central product of two groups such that I can see what it is step by step I would be very happy.

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The follwing is a step by step check of Pauli group G is the central product of subgroups H and K, where H and K are isomorphic to C4 and D4, respectively.

Definition of G, H, K:

G={{{-1,0},{0,-1}},{{-1,0},{0,1}},{{0,-1},{-1,0}},{{0,-1},{1,0}},{{0,-I},{-I,0}},{{0,-I},{I,0}},{{0,I},{-I,0}},{{0,I},{I,0}},{{0,1},{-1,0}},{{0,1},{1,0}},{{-I,0},{0,-I}},{{-I,0},{0,I}},{{I,0},{0,-I}},{{I,0},{0,I}},{{1,0},{0,-1}},{{1,0},{0,1}}};
H={{{1,0},{0,1}},{{I,0},{0,I}},{{-1,0},{0,-1}},{{-I,0},{0,-I}}};
K={{{1,0},{0,1}},{{0,1},{1,0}},{{0,-I},{I,0}},{{I,0},{0,-I}},{{0,-1},{-1,0}},{{-1,0},{0,-1}},{{-I,0},{0,I}},{{0,I},{-I,0}}};

MatrixForm /@ G

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MatrixForm /@ H

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MatrixForm /@ K

enter image description here

{SubsetQ[G, H], SubsetQ[G, K]}
(*{True, True}*)

A tool which generates the multiplication table of a group given as a list of matrices:

ClearAll[mymulttab];
 multtab[G_List]:=Module[{n=Length[G]},Grid[Table[Position[G,G[[i]].G[[j]]][[1,1]],{i,n},{j,n}]]];

multtab[H]

enter image description here

Grid[GroupMultiplicationTable[CyclicGroup[4]]]

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% === %%
(* True *)

multtab[K]

enter image description here

Grid[GroupMultiplicationTable[DihedralGroup[4]]]

enter image description here

% === %%
(* True *)

So, H is isomorphic to C_4 and K is isomorphic to D_4.

From now on, let us check the condions 1 and 2 of "internal central product" (of Wikipedia page cited in the comment).

Condition 1 : G is generated by H and K.

Union[G] === Union[Flatten[Outer[Dot, H, K, 1], 1]]
(* True *)

Condition 2 : Every element of H commutes with every element of K.

Union[Flatten[Table[y.x == x.y, {x, H}, {y, K}]]]
(* {True} *)
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  • $\begingroup$ @Kato This is direct and clear as a bell! Thank you. $\endgroup$ Commented Jun 24 at 2:05
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    $\begingroup$ @Dukes You're welcome. As for the choice of H and K, some trial and error was needed to make the multiplication tables match exactly :-) $\endgroup$
    – A. Kato
    Commented Jun 24 at 11:18

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