The congruence subgroup problem for a finitely generated group $Gamma$ and
$Gleq Aut(Gamma)$ asks whether the map $hat{G} o Aut(hat{Gamma})$ is
injective, or more generally, what is its kernel $Cleft(G,Gamma
ight)$? Here
$hat{X}$ denotes the profinite completion of $X$. In the case $G=Aut(Gamma)$
we denote $Cleft(Gamma
ight)=Cleft(Aut(Gamma),Gamma
ight)$.
Let $Gamma$ be a finitely generated group,
$ar{Gamma}=Gamma/[Gamma,Gamma]$, and
$Gamma^{*}=ar{Gamma}/tor(ar{Gamma})congmathbb{Z}^{(d)}$. Denote
$Aut^{*}(Gamma)= extrm{Im}(Aut(Gamma) o Aut(Gamma^{*}))leq
GL_{d}(mathbb{Z})$. In this paper we show that when $Gamma$ is nilpotent,
there is a canonical isomorphism $Cleft(Gamma
ight)simeq
C(Aut^{*}(Gamma),Gamma^{*})$. In other words, $Cleft(Gamma
ight)$ is
completely determined by the solution to the classical congruence subgroup
problem for the arithmetic group $Aut^{*}(Gamma)$.
In particular, in the case where $Gamma=Psi_{n,c}$ is a finitely generated
free nilpotent group of class $c$ on $n$ elements, we get that
$C(Psi_{n,c})=C(mathbb{Z}^{(n)})={e}$ whenever $ngeq3$, and
$C(Psi_{2,c})=C(mathbb{Z}^{(2)})=hat{F}_{omega}$ = the free profinite group
on countable number of generators.