A Problem in Group Theory solved by Computer Algebra

I shall report on how a problem in finite group theory could be translated to algebraic geometry and partially solved using computer algebra.

If G is a group and x, y ∈ G, we inductively define

e_{1}(x, y) := x^{−2} · y^{−1} · x, e_{n+1}(x, y) := [x · e_{n}(x, y) · x^{−1}, y · e_{n}(x, y) · y^{−1}] ,

where the commutator of g, h ∈ G is defined by [g, h] := ghg^{−1}h^{−1}.

Theorem (T. Bandman, G.-M. Greuel, F. Grunewald, B. Kunyavsky G. Pfister, E. Plotkin)

A finite Group G is solvable if and only if there is an n ∈ ℕ such that e_{n}(x, y) = 1 for all x, y ∈ G.

Recall that the minimal finite non–solvable groups G (that is, all subgroups of G are solvable) were classified by J. Thompson:

1. PSL(2, p), p a prime number, p = 5 or p > 5 and p = ±2 mod 5.

2. PSL(2, 2^{n}), n ≥ 2, a prime number.

3. PSL(2, 3^{n}), n odd, a prime number.

4. PSL(3, 3).

5. Sz(2^{n}), n odd.

To prove the theorem it is enough to show that for each group G in Thompson’s list we have x, y ∈ G with e_{1}(x, y) = e_{2}(x, y) and y ≠ x^{−1}. This implies 1 ≠ e_{1}(x, y) = e_{n}(x, y) for all n.

Here is the idea on how to prove the theorem for the group PSL(2,q). The case PSL(3,3) is easy but the case S_{z}(2^{n}) is much more difficult. Proposition If q = p^{k} for a prime p and q ≠ 2, 3, then there are x, y in PSL(2, 𝔽_{q}) with y ≠ x^{−1} and e_{1}(x, y) = e_{2}(x, y).

The proof will use some explicit computations with the following matrices. Let R be a commutative ring with identity and define

x(t) := (\begin{matrix}t & −1 \\ 1 & 0 \end{matrix}) , y(b, c) := (\begin{matrix}1 & b \\ c & 1 + bc \end{matrix}) ∈ SL(2, R)

for t, b, c ∈ R.

Let I = ⟨n_{1}, n_{2}, n_{3}, n_{4}⟩ ⊆ ℤ[b, c, t] be the ideal generated by the entries of the matrix e_{1}(x, y)− e_{2}(x, y).

To prove the above proposition, it is enough to prove the following

Proposition The variety V^{(q)} = V (I𝔽_{q}[b, c, t]) ⊂ 𝔽_{q}^{3} induced by I is not empty. We are going to apply the theorem of Hasse–Weil as generalised by Aubry and Perret to singular curves, and using the fact that the affine curve C has, at most, deg(\overline{C}) rational points less than the projective closure \overline{C}.

Proposition Let C ⊆ 𝔸^{n} be an absolutely irreducible affine curve defined over the finite field 𝔽_{q} and \overline{C} ⊂ ¶^{n} the projective closure, then the number of 𝔽_{q}–rational points of C is at least q + 1 − 2p_{a}\sqrt{q} − d with d = the degree and p_{a} the arithmetic genus of \overline{C}.

Note that the Hilbert polynomial of \overline{C}, H(t) = dt − p_{a} + 1, can be computed from the homogeneous ideal I_{h} of \overline{C}, hence we can compute d and p_{a} without any knowledge about the singularities of \overline{C}. In our case, we obtain H(t) = 10d − 11.

From the Hilbert polynomial, we read the degree d = 10 and the arithmetic genus p_{a} = 12. Using the proposition, we obtain:

#V^{(q)} ≥ q + 1 − 24\sqrt{q} − 10.

This implies that V^{(q)} is not empty if q > 593.

For the remaining cases, we check explicitly by computer that V^{(q)} is not empty.