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Trying to prove:

If $G$ is a torsion-free nilpotent group and $H \leq G$ has finite index, then the nilpotence classes of G and H are equal.

Obviously the class of H is at most that of G by taking intersections. I also know every factor group of upper central series for G is a torsion-free abelian group, and it seems intuitively sensible that H having finite index means it somehow preserves the "infiniteness" from the larger group.

I feel like there's some small bridge I'm missing between the index being finite and, maybe, the normal series of intersections of H with the central series of G not having any trivial factors. I'd appreciate a small hint to which direction to look or just a statement of what theorem I should be using; I have the nagging feeling it's something I should already know well and am just forgetting.

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    Letting $\gamma_k(G)$ be the lower central series, I think you can show by induction that $\gamma_{i+1}(G)\gamma_i(H)$ has finite index in $\gamma_i(G)$ for all $i$, which implies the result. I haven't got time to write down details right now! – 2017-01-28

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$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$This is a very slight variation on the suggestion of Derek Holt.

Consider the core $K$ of $H$, that is, the intersection of all conjugates of $H$. This has also finite index in $G$, and it is normal in $G$. We will prove that $K$ has the same nilpotence class as $G$, which implies the result for $H$.

Let us prove that for all $i$ we have that $\gamma_{i}(G)/\gamma_{i}(K) \gamma_{i+1}(G)$ has finite exponent. This is true by assumption if $i = 1$, as $G/K$ is finite, so take $m_{1} = \Size{G/K}$. Suppose $\gamma_{i}(G)/\gamma_{i}(K) \gamma_{i+1}(G)$ has finite exponent $m_{i}$, let $g \in \gamma_{i}(G)$, $h \in G$. Then $$ [g, h]^{m_{i} m_{1}} \in [g^{m_{i}}, h^{m_{1}}] \gamma_{i+2}(G) \le [\gamma_{i}(K), K] \gamma_{i+2}(G) = \gamma_{i+1}(K) \gamma_{i+2}(G). $$

So if $G$ has class $n$ and $K$ has class less than $n$, we have that $$ \gamma_{n-1}(G)/\gamma_{n}(K) \gamma_{n+1}(G) \cong \gamma_{n-1}(G) $$ has finite exponent, and thus is trivial, a contradiction.

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    Working through some of this, I think my biggest point of confusion is why you can freely change back and forth between $\gamma_i(K)$ and $\gamma_i(K) \gamma_{i+1}(G)$, which appear to be possibly very different groups. It's clear to me why the latter has finite index for i = 1, but not why we can say $g^{m_i}$ is in just $\gamma_i(K)$ instead of being possibly somewhere else in the product. – 2017-01-28
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    @Xindaris, thanks for the accepting this answer - I probably skipped some sensible step, please feel free to ask for more clarifications.As to your specific question, $g^{m_{i}} \in \gamma_{i}(K) \gamma_{i+1}(G)$, so commuting with something from $K$ will produce an element in $\gamma_{i+1}(K) \gamma_{i+2}(G)$. Besides that, I had failed to write a $\gamma_{i+1}(G)$ in one of the steps, hence your confusion, which is entirely my fault. – 2017-01-28