Suppose we have a nilpotent group $G$ that is generated by a subgroup $H$ and a single element $g$ of finite order. Is $H$ necessarily of finite index in $G$? What can be said of this index in any case?
Some context: The question is clear if $G$ is abelian (in which case the index is the order of $g$ in the quotient $G/H$). If $G$ is merely solvable (even if $[G,G]$ is already abelian), the index can be infinite as can for instance be observed on the infinite dihedral group $D_\infty$. For $G$ nilpotent, I do not know what happens, but strongly suspect that the index should always be finite. Note however that the index may well be much larger than the order of $g$: The dihedral group $D_{2^n}$ is generated by two reflections and thereby achieves an index of $2^n$ in this situation, compared to the order 2 of the generators.
If it helps, feel free to assume that $G$ is itself periodic.
Best Answer
The answer in general is no, even in nilpotency of class $2$.
Let $H$ be a direct sum of countably many copies of a cyclic group, say a cyclic group of order $p$, with the cyclic groups generated by elements $\{x_n\}_{n\in\mathbb{N}}$. Now let $G$ be generated by $H$ and $g$ and subject to the relations $$g^p=[[x_n,g],g] = [[x_n,g],x_m]=1\qquad n,m=1,2,3,\ldots$$ ($G$ is the $2$-nilpotent product of $H$ and $\langle g\rangle$; this example is torsion, by the way, in fact, if $p$ is odd it is of exponent $p$).
Then $G$ is nilpotent of class $2$, as its commutator subgroup is central. Moreover, every element of $G$ can be written uniquely in the form $$g^{\alpha}\prod_{i=1}^{\infty}x_i^{a_i}\prod_{j=1}^{\infty}[x_j,g]^{b_j}$$ where $0\leq \alpha,a_i,b_j\lt p$, and almost all $a_i$ and almost all $b_j$ are zero. Identifying such an element with the ordered triple $(\alpha,(a_i),(b_j))$, the multiplication in $G$ is given by $$\Bigl(\alpha,(a_i),(b_j)\Bigr)\Bigl(\beta,(c_i),(d_j)\Bigr) = \Bigl( \alpha+\beta, (a_i+c_i), (b_j+d_j+\beta a_j)\Bigr),$$ with addition modulo $p$, and $$\Bigl( \alpha,(a_i),(b_j)\Bigr)^{-1} = \Bigl( -\alpha,(-a_i), (a_j\alpha-b_j)\Bigr).$$
Then $H$ is of infinite index in $G$, since for example, $(0,(0),(b_j))$ and $(0,(0),(c_j))$ lie in the same coset of $H$ if and only if $(b_j)_{j\in\mathbb{N}}=(c_j)_{j\in\mathbb{N}}$.
The result would hold in class $2$ if $H$ is finitely generated, because the commutator subgroup is generated by elements of the form $[h,g]$ where $h$ ranges over a generating set of $H$, and each is of exponent the order of $g$; so the result holds in $H\amalg^{\mathfrak{N_2}}\langle g\rangle$ and hence in any $2$-nilpotent group generated by $H$ and $g$. I'm pretty sure that it would hold for any nilpotency class if $H$ is finitely generated, via a similar argument (again, the terms of the lower central series would be generated by basic commutators on a generating set for $H$ and $g$, and each of them would be of finite order bounded by a function in the order of $g$ and the nilpotency class). It also holds if $H$ is normal (which is what is "really" behind your argument for the abelian case), or more generally if $H$ is of finite index in its normal closure.