Suppose $n$ is a positive integer greater than 2, and $F$ is an arbitrary field with at least 4 elements.


Denote $\text{GL}\_n(F)$ the general linear group in the usual sense and $U\_n(F)$ the unipotent group in the sense that it consists of $n\times n$ upper-triangular matrices with 1's on the diagonal.


The inclusion $U\_n(F)\hookrightarrow\text{GL}\_n(F)$ induces a (unstable) homology homomorphism $f\_k: H\_k(U\_n(F);\mathbb{Z})\to H\_k(\text{GL}\_n(F);\mathbb{Z})$ for any positive integer $k$.


And there is another map, a stable homology map: $g\_k: H\_k(U\_n(F);\mathbb{Z})\to H\_k(\text{GL}(F);\mathbb{Z})$.


How can I detect the image of $f\_k$ (or $g\_k$)? Is $\text{im}(f\_k)$ or $\text{im}(g\_k)$ always zero?


Background: In Suslin's paper [1] (Sublemma 4.4.2), he asserts that when $k=n=\text{char}(F)=3$ then $\text{im}(f\_k)=\text{im}(g\_k)=0$ with a reference directed to [2]. But from [2] it's quite unclear to me how to deduce the assertion made in [1].


[1] A. Suslin, K3 of a field, and the Bloch group (Russian), Trudy Mat. Inst. Steklov
183 (1990), 180-199. English transl. in Proc. Steklov Inst. Math. (1991), 217-239.


[2] A. Suslin, [Stability in algebraic K-theory](https://link.springer.com/chapter/10.1007/BFb0062181), pp. 304-333 in Lecture Notes in Math.
966, Springer-Verlag, 1982.



How can I detect the homology image of a unipotent group in the general linear group?
Suppose first that $F$ is a finite field of characteristic $p$. Then $U\_n(F)$ is a Sylow $p$-subgroup of $GL\_n(F)$, and so using the transfer in group homology one sees that the image of $f\_k$ (for $k>0$) is precisely the $p$-torsion in $H\_k(GL\_n(F);\mathbb{Z})$. Unstably, this is not well understood, but it cannot be trivial by general principles in group cohomology (see the top answer to [this MO question](https://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree)). But it follows that the image of $g\_k$ is contain in the $p$-torsion of $H\_k(GL(F);\mathbb{Z})$, and this is completely understood by Quillen's calculations: it is trivial.


Suppose now that the field $F$ is infinite; it is then a "ring with many units" in the sense of Nesterenko--Suslin (*Homology of the full linear group over a local ring, and Milnor's K-theory*), and we may apply the results of that paper. There is a group $UT\_n(F)$ of all upper triangular matrices (not necessarily having 1's on the diagonal), and a split extension
$$1 \to U\_n(F) \overset{i}\to UT\_n(F) \overset{q}\to D\_n(F) \to 1$$
where $D\_n(F)$ are the diagonal matrices. If follows by iteratedly applying Theorem 1.11 of Nesterenko--Suslin that the map $q$ induces an isomorphism in integral homology, but then $i$ induces the trivial map in integral homology. As the inclusion of $U\_n(F)$ into $GL\_n(F)$ factors over $UT\_n(F)$, it follows that $f\_k$ (and hence $g\_k$) is trivial.


