Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$

Question

I’m wondering what is known about the cohomology of $\operatorname{GL}_3(\mathbb{Z}_2)$, the general linear group of $3\times3$ matrices over the finite field $\mathbb{Z}_2$. There are results in Chapter I of Knudson’s “Homology of Linear Groups” as well as Chapter VII of Adem and Milgram’s “Cohomology of Finite Groups” that consider the cohomology of $\operatorname{GL}_n(\mathbb{F}_p)$, but not this specific case.

Update: It seems that Adem and Milgram does the trick once we identify $\operatorname{GL}(3,2) =\operatorname{SL}(3,2)$. The relevant lemma says that $$H^*(\operatorname{SL}(3,2)) \oplus H^*(D_8) \cong H^*(S_4)\oplus H^*(S_4). $$

Answer 1

As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), you can use GAP, specifically the hap package. The following code computes $H^i(G, \mathbf{Z})$ for $i \leq 10$:

LoadPackage("hap"); 
G = GL(3,2); 
List([1..10], x->GroupCohomology(G, x));

This will output:

[ [  ], [  ], [ 2 ], [ 4, 3 ], [  ], [ 2, 2, 7 ], [ 2 ], [ 4, 3 ], [ 2, 2 ], [ 2, 2 ] ]

which is a list of the abelian invariants of the cohomology groups $H^i(G, \mathbf{Z})$ for $1 \leq i \leq 10$, i.e. the first ten integral cohomology groups are $1, 1, C_2, C_4 \times C_3, 1, C_2 \times C_2 \times C_7, C_2, C_4 \times C_3, C_2 \times C_2$ and $C_2 \times C_2$ (for readability I here use $C_n = \mathbf{Z}/n\mathbf{Z}$).

If you want to compute the mod $p$ cohomology, and don't feel like using the above with the universal coefficient theorem by hand, you can compute this just as easily by using GroupCohomology(G, x, p). When $p=2$, the sequence of exponents of the first ten cohomology groups (together with $H^0(G, \mathbf{Z}/2\mathbf{Z})$) is $$ 1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3 $$ e.g., $H^3(G, \mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})^2$. This sequence (of course) agrees with the generating function in @DaveBenson's answer, which expands as: $$ \frac{1+t^3}{(1-t^2)(1-t^3)} = 1 + \quad + t^2 + 2 t^3 + t^4 + 2 t^5 + 3 t^6 + 2 t^7 + 3 t^8 + 4 t^9 + 3 t^{10} + ... $$

Answer 2

For the mod $2$ cohomology, see Section 11 of Benson and Carlson, “Diagrammatic methods for modular representations and cohomology” Comm in Alg 15 (1987), 53-121 for the ring structure. If you want to know about the $A_\infty$ structure, and much more information besides, see Benson - Classifying spaces of finite groups of tame representation type. At odd primes, the group has cyclic Sylow subgroup, so the cohomology is easy.

In the mod 2 cohomology, as given in the first reference above, there's a generator in degree 2, two generators in degree 3, and one relation, saying that the product of the two degree 3 generators is equal to zero. This gives a generating function for the dimensions of $\frac{1+t^3}{(1-t^2)(1-t^3)}$.