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Add the link to cited Shalen-Wagreich article
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Luc Guyot
Luc Guyot
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Does the Shalen-Wagreich propositionlemma holds for non-symmetric generating sets?

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 andShalen-Wagreich's Lemma 3.4 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?

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 lemma holds for non-symmetric generating sets?

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's Lemma 3.4 and $k=2d-1$. But I wonder whether it still holds when $S$ is non-symmetric? Or is there a counterexample?

edited tags, fixed English
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YCor
YCor
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Whether Does the Shalen-Wagreich proposition holds for non-symmetric generating sets?

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 dependendingdepending 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?

Whether the Shalen-Wagreich proposition holds for non-symmetric generating sets?

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 dependending 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?

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?

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dennis
dennis
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Whether the Shalen-Wagreich proposition holds for non-symmetric generating sets?

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 dependending 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?