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 $x$, $y$, $z$?
Generators of $GL(2, \mathbb{F}_3)$
2 Answers
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).
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0Thanks, but I don't know what is GAP. Is it possible to see that we have two different groups by comparing some invariants of the groups such as number of elements of fixed order, automorphisms or any other invariants. – 2011-11-29
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0@Alex: I added a link to gap and a reasonably clear difference with their elements of order 2: the "other" group has a unique element of order 2, while GL(2,3) has 13 elements of order 2. – 2011-11-29
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0A 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
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 $y = \left( \begin{array}{clcr} 2 & 2\\1 & 0 \end{array} \right) .$