$PSL(2,7)$ as a (transitive) subgroup of $AL(8)$

Question

N. Elkies' page http://www.math.harvard.edu/~elkies/trinomial.html ends with an information about octic trinomials "whose Galois group is contained in $G_{1344}$".

One of reported trinomials, $x^8+324x+567$, "has a smaller Galois group, which embeds as a transitive subgroup into $G_{1344}$ (...) This Galois group is isomorphic with $G_{168}$, acting on 8 letters via the other guise of that group, as $PSL_{2}(\mathbb{Z}/7\mathbb{Z})$".

Using GAP one can check that indeed mentioned trinomial has Galois group $PSL(2,7)$:

gap> x:=Indeterminate(Rationals, "x");;
gap> GaloisType(x^8+324*x+567);
37
gap> t37:=TransitiveGroup(8,37);
L(8)=PSL(2,7)
gap> Size(t37);
168

In the context of the above web page it is surprising for me that $PSL(2,7)$ is not a subgroup of $G_{1344}$:

gap> t48:=TransitiveGroup(8,48);
E(8):L_7=AL(8)
gap> Size(t48);
1344
gap> IsSubgroup(t48, t37);
false

Moreover, there is another Galois group $G_{168}$, $C_2^3:(C_7: C_3)$, which is a subgroup of $G_{1344}$:

gap> t36:=TransitiveGroup(8,36);
E(8):F_21
gap> Size(t36);
168
gap> IsSubgroup(t48, t36);
true

Question: How should I understand this contradiction/"contradiction"?

Answer 1

TransitiveGroup will give you one representative from a conjugacy class of subgroups. For two subgroups $A$, $B$ in the library it is not guaranteed (in fact not guaranteeable) that if a conjugate of $A$ is contained in $B$, then automatically $A$ is contained in $B$.

To test for such inclusions, you will need to either calculate subgroups of $B$ and find $A$ there, or search for IsomorphicSubgroups (which finds monomorphism into the larger group):

gap> homs:=IsomorphicSubgroups(t48,t37);
[ [ (1,8,2)(3,7,6), (1,2)(3,5)(4,7)(6,8) ] -> [ (2,4,7)(3,6,8), (4,5)(6,7) ],
  [ (1,8,2)(3,7,6), (1,2)(3,5)(4,7)(6,8) ] ->
  [ (2,4,7)(3,6,8), (1,3)(2,8)(4,7)(5,6) ] ]
gap> u:=Image(homs[1],t37); 
Group([ (2,4,7)(3,6,8), (4,5)(6,7) ])
gap> NrMovedPoints(u); TransitiveIdentification(u);
7
5

So you see that the subgroup isomorphic to $L_2(7)$ is in fact TransitiveGroup(7,5). (the groups are isomorphic but not permutation isomorphic.)

Answer 2

The group $G_{168}$ is isomorphic to $PSL(2,7)$, and it is a subgroup of $G_{1344}$. Elkies writes "This Galois group is isomorphic with $G_{168}$, acting on $8$ letters via the other guise of that group, as $PSL_2(\mathbb{Z}/7\mathbb{Z})$." So you could understand the "contradiction" reviewing his word "the other guise".