either in $E_n(A)$ (the subgroup generated by the elementary matrices) or in $GL_n(A)$ itself. Indeed, the set of orbits $Um_n(A)/GL_n(A)$ detects the stably free non free modules of rank $n-1$ as follows:

Let $v=\uni na$ be a unimodular row and let $P(v)$ be the projective module defined by the exact sequence $$\xymatrix@C=2em{0\ar[r] & P(v)\ar[r] & A^n\ar[r]^-{v} & A\ar[r] & 0.}$$ It is easily seen that the following proposition holds:

\[free\] The modules $P(v)$ and $P(v^\prime)$ are isomorphic if and only if there is a matrix $G\in GL_n(A)$ such that $vG=v^\prime$.

This proposition also shows that the set $Um_n(A)/GL_n(A)$ is quite hard to compute in general. If one restricts to $E_n(A)$, the situation is slightly better, as shown by the next result due to W. van der Kallen ([@vdk Theorem 4.1]):

Let $A$ be a ring of dimension $d$. If $n\geq (d+4)/2$ then $Um_n(A)/E_n(A)$ has the structure of an abelian group.

Of course, this doesn’t imply that $Um_n(A)/E_n(A)$ becomes suddenly easy to compute. However, in the special situation when $A$ is a smooth algebra of dimension $d$ over a field of characteristic different from $2$, it is shown in [@Fa4] that $Um_{d+1}(A)/E_{d+1}(A)$ (same $d$) has a cohomological interpretation (\[loc. cit., Theorem 