$K_S$ modulo $K_S^n$, where $K_S$ is the group of $S$-units Let $K$ be a field containing the $n$th roots of unity, $S$ a finite set of places containing all the archimedean ones, and $K_S$ the group of $S$-units, i.e. those $x \in K^{\ast}$ which are units at all places outside $S$.  It is a consequence of the unit theorem that there exist $c_1, ... , c_{s-1} \in K_S$ such that every $x \in K_S$ can be uniquely expressed as $$\zeta c_1^{m_1} \cdots c_{s-1}^{m_{s-1}}$$ where $m_i$ are integers and $\zeta$ is a root of unity in $K$.  Serge Lang claims (Algebraic Number Theory, page 216):

$K_S$ modulo the $n$th roots of unity is a free abelian group on $s-1$ generators.

and uses this to conclude that $K_S/K_S^n$ has $n^s$ elements.  However, is his statement correct?  $K_S$ modulo the group of roots of unity in $K$ is free abelian of rank $s-1$.  There could be other roots of unity in $K$ besides $n$th roots of unity.

I don't think $K_S$ modulo the $n$th roots of unity is free of rank $s-1$ in general, but $[K_S : K_S^n] = n^s$ is still true.  By the unit theorem, we can write $K_S$ as an internal direct sum $H \times T$, where $H$ is the group of units and $T$ is a free $\mathbb{Z}$-module of rank $s-1$.  Here $T/T^n$ is isomorphic to $\bigoplus\limits_{i=1}^{s-1} \mathbb{Z}/n\mathbb{Z}$, so it has $n^{s-1}$ elements.  Since $H$ contains the $n$th roots of unity, $n$ divides $H$.  Since $H$ is cyclic, $H^n$ has $\frac{1}{n} |H|$ elements, so $H/nH$ has $n$ elements.  Hence $$K_S/K_S^n \cong \frac{H \oplus T}{H^n \oplus T^n} \cong H/H^n \oplus T/T^n$$ has $n^s$ elements.  
