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I came across the statement that for $GL(2, \mathbb{F}_3)$ we can choose generators $x$, $y$, $z$, that satisfy relations $x^4=y^3=z^2=xyz$, $(xyz)^2=1$. What matrices can I take for $x$, $y$, $z$?

2 Answers 2

8

The statement is not true, though it is close. You can check this using the small groups library from within the computer algebra system GAP:

gap> f := FreeGroup( "x", "y", "z" );; gap> g := f / ParseRelators( GeneratorsOfGroup(f), > "x^4 = y^3 = z^2 = xyz, (xyz)^2 = 1" );; gap> IdGroup( g ); [ 48, 28 ] gap> IdGroup( GL(2,3) ); [ 48, 29 ] gap> StructureDescription(g/Intersection(DerivedSubgroup(g),Center(g))); "S4" 

Thus the group defined by that presentation is a group of order 48 that is a double cover of S4 ≅ PGL(2,3), but it is the "other" Schur cover, not GL(2,3). One important difference between these groups is their 2-local structure. The Sylow 2-subgroup of GL(2,3) is a quasi-dihedral group of order 16, while the Sylow 2-subgroup of the "other" group is quaternion of order 16. This might be easier to see at the element level: GL(2,3) has two different conjugacy classes of involutions (elements of order 2), while the "other" group has a unique class of involutions (both of these statements are also true for their Sylow 2-subgroups).

If you want matrices for this "other" group, you can again use GAP:

# In GAP 4.5: gap> g := SchurCoverOfSymmetricGroup(4,3,1);; # In GAP 4.4: gap> g := Group([ [ [ 0*Z(3), Z(3) ], [ Z(3)^0, Z(3^2)^6 ] ], > [ [ Z(3^2)^2, 0*Z(3) ], [ 0*Z(3), Z(3^2)^6 ] ] ]);; # In either: gap> xyz := Filtered( Tuples(g,3), >  function(xyz) >    local x,y,z; >    x:=xyz[1]; >    y:=xyz[2]; >    z:=xyz[3]; >    return x^4 = y^3 and y^3 = z^2 and z^2 = x*y*z and (x*y*z)^2 = x^0 >    and Subgroup( g, xyz ) = g; >  end );; gap> orb := Orbits( AutomorphismGroup(g), xyz, OnTuples );; gap> Size(orb); 1 

so there is only one such generating set up to isomorphism. We can display it in GAP:

gap> Display(orb[1][1][1]); gap> Display(orb[1][1][2]); gap> Display(orb[1][1][3]); 

But TeX is a bit nicer:

$ x = \left(\begin{array}{rr}0&1\\-1&\sqrt{2}\end{array}\right), \quad y = \left(\begin{array}{rr}1&\sqrt{-1}\\\sqrt{-1}&0\end{array}\right), \quad z = \left(\begin{array}{rr}\sqrt{-1}&0\\-1+\sqrt{-2}&-\sqrt{-1}\end{array}\right), \quad $

where √2 ≡ −√−1 mod 3, and √−2 ≡ 1 mod 3.

This defines a subgroup of GL(2,9) that is isoclinic but not isomorphic to GL(2,3). Multiplying x and z by √−1 gives an isomorphic copy of GL(2,3).

  • 0
    A Sylow 2-subgroup of your $\langle x,y,z:x^4=y^3=z^2=xyz, (xyz)^2=1\rangle$ is generated by $x$ and $yx^{-1}y$ and you can use the 2-dimensional representation in my answer to easily check that the subgroup is quaternion of order 16.2011-11-29
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I just remark that ${\rm GL}(2,3)$ has a presentation $\langle x,y,z : x^8 = y^3 = z^2 = (xyz) = [x^4,y ] = 1 \rangle.$ This presentation gives a double cover of $S_4,$ but one which contains more than one involution, since $z \neq x^4.$ You can take $z = \left( \begin{array}{crclcr} 1 & 0\\ 0 & 2 \end{array} \right)$ and $y = \left( \begin{array}{clcr} 2 & 2\\1 & 0 \end{array} \right) .$