The isomorphism GL_{4}(F_{2}) ≅ A_{8} (outline)

Noam D. Elkies

The simple groups A_{8} and GL_{4}(F_{2}) both have \frac{1}{2}8! = 15 · 14 · 12 · 8 = 20160 elements. One might then expect that they are isomorphic. This in fact turns out to be the case. One way to show this is to construct a 4-dimensional vector space V over F_{2} on which A_{7} acts nontrivially by linear transformations; this will embed A_{7} as an index-8 subgroup of GL_{4}(F_{2}), and thus a homomorphism of GL_{4}(F_{2}) into S_{8}. Since GL_{4}(F_{2}) is simple, this homomorphism will have trivial kernel, and image contained in A_{8}. It will thus yield an embedding of GL_{4}(F_{2}) into A_{8}, which must then be an isomorphism because the group orders coincide.

Let H ⊂ S_{7} be the automorphism group of the projective plane Π_{2}. We know that #H = 168. Moreover, H is simple, so H ⊂ A_{7}. Thus H has [A_{7} : H] = \frac{1}{2}7!/168 = 15 cosets in A_{7}. To these 15 we will add 0 to obtain V, and show that V has the structure of an F_{2}-vector space. That is, we will define for each P, Q ∈ V a natural element P + Q ∈ V and show that this addition law satisfies the axioms of an F_{2}-vector space.

The cosets of H in S_{7} are the 30 different ways to identify an unstructured 7-element set Σ with the points of Π_{2}. We’ll show that two of these fall in the same A_{7}-orbit if and only if they have exactly one line in common. (Otherwise they share either three lines or none, but we will not need this.) We do this by noting that given one Π_{2} structure P on Σ and one of its seven lines l, there are two other Π_{2} structures in the same A_{7}-orbit whose only common line with P is l, namely the images of P under g, g^{2} where g ∈ A_{7} is a 3-cycle permuting l. This is enough because it accounts for all 1 + 2 · 7 = 15 structures in the orbit of P. We now define P + Q for P, Q ≠ 0 as follows: if P = Q then of course P + Q = 0; else P, Q have a unique line l in common, and P + Q is the third Π_{2}-structure in the same A_{7} containing l. The only obstruction to verifying that this makes V into an F_{2}-vector space is checking that the identity (P + Q) + R = P + (Q + R) for distinct P, Q, R ∈ V − {0} without a common line.

Fortunately A_{7} acts (simply) transitively on the 15 · 14 · 12 = 2520 such triples (P, Q, R), so it is enough to check the identity on any one of them. To prove the transitivity it suffices to verify that if g ∈ A_{7} fixes each of P, Q, R then g is the identity. But then g fixes each of the lines common to two of P, Q, R, which are three 3-element subsets of Σ any two of which meet at a point. They thus separate Σ into singletons, and so g cannot move any element of Σ. [Note that it follows in particular that the action of A_{7} on V − {0} is doubly transitive.] Confirming that (P + Q) + R = P + (Q + R) for some appropriate P, Q, R is straightforward. This completes the identification of GL_{4}(F_{2}) with A_{8}.

Alternatively we could have obtained V directly as a representation of A_{8}, by identifying the nonzero points with orbits of AGL_{3}(F_{2}) ⊂ A_{8}, i.e. with half of the 30 ways of arranging an 8-element set into a 3-(8,4,1) Steiner system. That makes it somewhat trickier to define the linear structure on V, though.

We can even recover S_{8} from GL_{4}(F_{2}). Recall that S_{8} = Aut(A_{8}), so we need only find an outer automorphism of GL_{4}(F_{2}). But we already know such an automorphism: the “contragredient” involution A ↔^{t}A^{−1}.