This question has been frustrating me for a while, because I'm pretty sure the solution is simple and I'm just not seeing it. I have found the claim (e.g., in Segal's Polycyclic Groups) that if $G$ is polycyclic and $H\leq G$, then $h(G) = h(H)$ if and only if $[G:H]<\infty$. The "if" direction is trivial, but I'm having trouble with the "only if" direction. (In Segal, this is left as an exercise.) If $H$ is subnormal in $G$, then the claim is easy. But I don't think a subgroup of a polycyclic group is necessarily subnormal, and I haven't managed to prove the claim if $H$ is not subnormal. Can anybody point me in the right direction?



Infinite index subgroup of polycyclic group has strictly lower Hirsch length?
Consider the series $A\_0 < A\_1 < ... < A\_n=G$ with cyclic factors. Suppose $G/A\_{n-1}$ is infinite (if it is finite, replace $G$ by $A\_{n-1}$). So $h(G)=h(A\_{n-1})+1$. Let $H$ be your subgroup of $G$. Consider its intersection $H'=H\cap A\_{n-1}$. Then $H'$ is normal in $H$ and $H/H'$ is cyclic. Assume $h(H)=h(G)$. Then $h(H')=h(A\_{n-1})$. By induction, $H'$ is of finite index in $A\_{n-1}$. Also note that $HA\_{n-1}$ is of finite index in $G$ (since every non-trivial subgroup of a cyclic group is of finite index and $H$ is not inside $A\_{n-1}$). We can as well assume that $HA\_{n-1}=G$ (we can always replace $G$ by its finite index subgroup containing $H$). But the representatives of the cosets of $H'=H\cap A\_{n-1}$ in $A\_{n-1}$ are representatives of cosets of $H$ in $HA\_{n-1}$, so $H$ has finite index in $G$. 


