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](https://doi.org/10.1007/978-3-0348-8338-2)” as well as Chapter VII of Adem and Milgram’s “[Cohomology of Finite Groups](https://doi.org/10.1007/978-3-662-06280-7)” 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
$$H^\*(\operatorname{SL}(3,2)) \oplus H^\*(D\_8) \cong H^\*(S\_4)\oplus H^\*(S\_4). $$



Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$
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](https://www.gap-system.org/index.html), 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:
$$
\frac{1+t^3}{(1-t^2)(1-t^3)} = 1 + \quad + t^2 + 2 t^3 + t^4 + 2 t^5 + 3 t^6 + 2 t^7 + 3 t^8 + 4 t^9 + 3 t^{10} + ...
$$


