Forwarded message from Ignat Soroko: ####################################
Dear GAP-Forum, Nicola Sottocornola, Jan Draisma, Volkmar Felsch and
Joachim Neubueser,
Some time ago Nicola Sottocornola asked:
I'd like to generate a group using two elements A and S: G = <A,S>. I
don't know if this group is finite. Someone can help me?
S = diag(1,1,-1,-1)
A = [0, 0, cos(s), -sin(s)] [-sin(t), cos(t), 0, 0] [cos(t), sin(t), 0, 0] [0, 0, sin(s), cos(s)] with cos(t) := 1/4 + 1/4 sqrt(5) sin(t) := 1/4 sqrt(2) sqrt(5 - sqrt(5)) cos(s):= 3/4 - 1/4 sqrt(5) sin(s):= 1/4 sqrt(2 + 6 sqrt(5))
Jan Draisma proposed an approach that had some subtle moments in it.
Volkmar Felsch and Joachim Neubueser in a letter of Fri 01, March 2002
offered an alternative approach, showing how GAP can handle
non-cyclotomic irrationalities. In my opinion, this approach should be
given in the GAP tutorial as a very instructive example.
I would like to note, that there is an easier way to prove using Maple
that the group G=<A,S> is infinite, but of course Maple doesn't have
in-built procedures for analysing the structure of this group, so we
have limited possibilities to investigate this group further.
Here is my idea.
If we enter matrices for A and S in Maple and then begin to
multiply them, we easily discover that A has order 6, A*S has
order 10, but A^2*S and A^3*S seem to have infinite order.
How one can prove that? Let's find the Jordan form of e.g. A^3*S.
As Maple told me, it equals
diag(1,-1,a,a_), where
a = 1/2*( sqrt(5)-1 + i * sqrt( 2*sqrt(5)-2 ) )
and a_ is the complex conjugate to a.
Now we must show that either a or a_ is not a root of unity. It
suffices to show that a is not a cyclotomic number. Some
computations show that the irreducible polynomial over
the rationals having numbers a and a_ as roots is
x^4+2*x^3-2*x^2+2*x+1
Calculating its Galois group over the rationals in Maple or
in GAP3:
gap> x:=X(Rationals);;x.name:="x";; gap> p:=x^4+2*x^3-2*x^2+2*x+1;;e:=AlgebraicExtension(p);; gap> TransitiveGroup(e.degree,GaloisType(e)); D(4) gap> IsAbelian(last); false
one finds out that it is isomorphic to the non-abelian dihedral
group of order 8, and it cannot be the quotient of Galois group
of any cyclotomic field because Gal(CF(m),Q)=(Z/mZ)^*, an abelian
group. This means that the numbers a and a_ cannot be embedded
into any cyclotomic field, hence they cannot be roots of unity.
This proves that A^3*S has infinite order and the group G is
in fact infinite.
Interestingly enough, the Jordan form of the matrix A turned out
to be diag(1,-1,E(3),E(3)^2), i.e. a symmetry of order 3 can be
given in terms of rotations expressed using sqrt(5)!
With best regards,
Ignat Soroko
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