Let $G$ be a group finitely generated by a set $S$ with $S=S^{-1}$. The word length of an element $g \in G$ (with respect to $S$) is defined as \begin{eqnarray} |g|_S:= \min \{n \in \mathbb{N} \mid s_1...s_n =g \text{ for some }s_1,...,s_n\in S\}. \end{eqnarray}
For a subgroup $H \leq G$, can we always find a finite generating set $T \subseteq H$ such that the restriction of $| \cdot |_S$ to $H$ coincides with the word length coming from $T$? If no, what about the case where $H$ is a (finite index) normal subgroup?
This is definitely not true. Consider the abelian group $\mathbb Z/2\mathbb Z\oplus\mathbb Z/2\mathbb Z$ with generators $S=\{(1, 0), (0, 1)\}$, then $|(1, 1)|_S=2$. But the subgroup generated by $(1, 1)$ has exactly one possible symmetric generating set i.e. $T=\{(1, 1)\}$ itself, and $|(1, 1)|_T = 1$. In this example, $H$ is normal and of finite index.
In general, it's just wrong to think the word metric as precise, but only the equivalence class based on quasi-isometry matters.