$H$. Since every subgroup of finite index contains a normal subgroup of finite index, it is enough to show that if $K$ is a normal subgroup of finite index in $H$ and $K \le H \le G$ with $H$ torsion free, then for every $k*H$-module $N$, $$\frac{\dim^{{\operatorname{Ore}}}_{k*K}\bigl({\operatorname{res}}_{k*H}^{k*K} N\bigr)}{[H:K]}
=
\dim^{{\operatorname{Ore}}}_{k*H}(N).
\label{Ore-dimension_and_restriction}$$ Let $T$ denote the set of non-zero-divisors of $k*H$ and write $S = (k*K) \cap T$. Note that $\Delta^+(K) = 1$, so $k*K$ is still a prime left Goldie ring and hence the ring $S^{-1}k*K$ exists. Then $S^{-1}k*H \cong (S^{-1}k*K)*[H/K]$ and there is a natural ring monomorphism $\theta\colon S^{-1}k*H \hookrightarrow T^{-1}k*H$. Since $S^{-1}k*K$ is a matrix ring over a skew field by Theorem \[thm:prime\_Goldie\_matrix\], we see that $(S^{-1}k*K)[H/K]$ is an Artinian ring, because $H/K$ is finite. But every element of $T$ is a non-zero-divisor in $(S^{-1}k*K)[H/K]$, and since every non-zero-divisor in an Artinian ring is invertible (compare [@Rowen(2008)]\*[Exercise 22 of Chapter 15 on p. 16]{}), we see that every element of $T$ is invertible in $S^{-1}k*H$ and we conclude that $\theta$ is onto and hence is an isomorphism. We deduce that $\dim_{S^{-1}k*K}(T^{-1}k*H) = [H:K]$ and that the natural map $S^{-1}N \to T^{-1}N$ induced by $s^{-1}n \mapsto
s^{-1}n$ is an 