0
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For example for the group $SL_2(\mathbb{F}_3)$ I get the following,

gap> IrreducibleRepresentations(SL(2,3));
[ CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ 1 ] ], [ [ 1 ] ] ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ E(3)^2 ] ], [ [ 1 ] ] ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ E(3) ] ], [ [ 1 ] ] ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ E(3), -E(3) ], [ 0, E(3)^2 ] ], [ [ E(3), 1 ], [ E(3), -E(3) ] ]
 ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ E(3)^2, E(3) ], [ 0, 1 ] ],
  [ [ -E(3)^2, -E(3) ], [ -E(3), E(3)^2 ] ] ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ E(3), E(3)^2 ], [ 0, 1 ] ], [ [ 0, 1 ], [ -1, 0 ] ]
 ], <action isomorphism> ),
  CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] ->
[ [ [ 0, 0, 1 ], [ 0, 1, 0 ], [ -1, -1, -1 ] ],
  [ [ 0, 1, 0 ], [ 1, 0, 0 ], [ -1, -1, -1 ] ] ], <action isomorphism> ) ]

It seems to me that it is specifying the two matrices in in each irrep of $SL_2(\mathbb{F}_3)$ which correspond to the two elements which it denotes as $(4,5,6)(7,9,8)$ and $(2,7,3,4)(5,8,9,6)$. But I don't know what is its scheme of denoting of denoting the group elements are? (what group elements do these two pairs of tuples correspond to?) And how to get the matrix for every group element from this?

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  • $\begingroup$ I don't know if this is correct, so I am not listing this as an answer. But...it appears to me that $(4, 5, 6)(7, 9, 8)$ and $(2, 7, 3, 4)(5, 8, 9, 6)$ are the generators of $SL_2(3)$, which it is somehow internally (and now externally) storing as a permutation group. I'd guess, therefore, that these permutations correspond to the generators of $SL_2(3)$ as matrices, which you can view if you call GeneratorsOfGroup(SL(2,3));. As slight confirmation, you can note that the group orders of these two matrices are 3 and 4: call List(GeneratorsOfGroup(SL(2,3)), Order);. $\endgroup$ – mathmandan Apr 29 '15 at 21:55
  • $\begingroup$ "GeneratorsOfGroup(SL(2,3))" is not a command on GAP $\endgroup$ – user6818 Apr 29 '15 at 21:57
  • $\begingroup$ List(GeneratorsOfGroup(SL(2,3))); [ [ [ Z(3)^0, Z(3)^0 ], [ 0*Z(3), Z(3)^0 ] ], [ [ 0*Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ] ] $\endgroup$ – user6818 Apr 29 '15 at 21:59
  • $\begingroup$ Are you sure? It works absolutely fine for me (without the quotation marks, and with a semicolon). I'm using GAP 4.6.2, so you may have a different version. $\endgroup$ – mathmandan Apr 29 '15 at 22:00
  • $\begingroup$ How does one map this above list to which of these two permutations it might correspond to? $\endgroup$ – user6818 Apr 29 '15 at 22:00
3
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What GAP returns is a list of homomorphisms, i.e. the actual irreducible representations. Each of these homomorphisms is a bit more complicated, as it actually factors from SL(2,3) through an isomorphic permutation group. What you can do is to take any such homomorphism, and evaluate it at arbitrary elements of SL(2,3).

gap> G:=SL(2,3);;
gap> irr:=IrreducibleRepresentations(G);
[ CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] -> 
    [ [ [ 1 ] ], [ [ 1 ] ] ], <action isomorphism> ), [...]

gap> rep:=irr[6];
CompositionMapping( [ (4,5,6)(7,9,8), (2,7,3,4)(5,8,9,6) ] -> 
 [ [ [ -E(3)^2, -E(3) ], [ 1, 0 ] ], [ [ 0, 1 ], [ -1, 0 ] ] ],
 <action isomorphism> )
gap> mat:=[[1,0],[1,1]]*One(GF(3));
[ [ Z(3)^0, 0*Z(3) ], [ Z(3)^0, Z(3)^0 ] ]
gap> Image(rep,mat);
[ [ -E(3), E(3)^2 ], [ -1, 0 ] ]

i.e. this group element is mapped to $\left(\begin{array}{cc}-\zeta&\zeta^2\\ -1&0\end{array}\right)$ where $\zeta=e^{\frac{2\pi i}{3}}$.

If you want to see what happens with generators under all representations, you could use:

gap> gens:=GeneratorsOfGroup(G);
[ [ [ Z(3)^0, Z(3)^0 ], [ 0*Z(3), Z(3)^0 ] ], 
  [ [ 0*Z(3), Z(3)^0 ], [ Z(3), 0*Z(3) ] ] ]
gap> List(irr,r->List(gens,g->Image(r,g)));
[ [ [ [ 1 ] ], [ [ 1 ] ] ], [ [ [ E(3)^2 ] ], [ [ 1 ] ] ], 
  [ [ [ E(3) ] ], [ [ 1 ] ] ], 
  [ [ [ E(3), -E(3)^2 ], [ 0, E(3)^2 ] ], 
      [ [ -E(3)^2, 1 ], [ E(3)^2, E(3)^2 ] ] ], 
  [ [ [ E(3)^2, -E(3) ], [ 0, 1 ] ], [ [ E(3), E(3)^2 ], [ E(3)^2, -E(3) ] ] ]
    , [ [ [ -E(3)^2, -E(3) ], [ 1, 0 ] ], [ [ 0, 1 ], [ -1, 0 ] ] ], 
  [ [ [ -1, -1, -1 ], [ 0, 1, 0 ], [ 1, 0, 0 ] ], 
      [ [ 0, 1, 0 ], [ 1, 0, 0 ], [ -1, -1, -1 ] ] ] ]
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  • $\begingroup$ What is the isomorphism map that GAP uses from elements of SL_2(F_p) to S_p^2 ? It clearly has one in mind when it chose to represent the generators as permutations! $\endgroup$ – user6818 Apr 30 '15 at 3:46
  • $\begingroup$ The Isomorphism is stored as NiceMonomorphism(G), though generically one also could use IsomorphismPermGroup. In this particular case it is simply the action on nonzero vectors. $\endgroup$ – ahulpke Apr 30 '15 at 13:29
  • $\begingroup$ This is not clear. What is this E(3) ? If 0 is occurring in the generators that GAP is using then it is having F_3 in the additive representation which has elements (0,1,-1) and there is no clear meaning to E(3). What is permutation representation that GAP is using to write its generators in a cycle decomposed notation? $\endgroup$ – user6818 May 1 '15 at 1:08
  • $\begingroup$ @user6818 E(3) is explained in my answer (it is $\zeta$), and the permutation action is on vectors. I have difficulties parsing the rest of your comment. $\endgroup$ – ahulpke May 1 '15 at 1:21
  • $\begingroup$ This interpretation of E(3) or Z(3) doesn't make sense. Firstly is E(3) the same as Z(3) ? I think there are two natural ways to think of F_p, - as the set {0,1,2,3..,p-1} - or as the set {0,1,x,x^2,...,x^{p-2}} where x is the (p-1)^th primitive root of unity Now what is GAP thinking when it says that [[0,1],[Z(3),0]] is one its generators of SL_2(F_3)? F_3 can either be represented in {0,1,-1} or as {0,1,2} What is Z(3) or E(3) ? $\endgroup$ – user6818 May 1 '15 at 17:12

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