Let $G$ be a group and $H$ a subgroup with finite index $d$. Then for any finite generating set $S$ of $G$, does $S^{\le k}$ contain a generating set of $H$ where $k$ is a constant depending only on $d$? When $S$ is symmetric, i.e. $S=S^{-1}$, the conclusion holds by Shalen-Wagreich proposition and $k=2d-1$. But I wonder whether it still holds when $S$ is non-symmetric? Or is there a counterexample?
Does the Shalen-Wagreich proposition holds for non-symmetric generating sets?
dennis
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