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Let $G$ be a group, $S$ its finite set of generators. Denote $G^+$ the subsemigroup of $G$ generated by $S$. Let $g\in G$, $H<G$ the subgroup generated by $g$ and let $g_1,...,g_k\in G$ be a finite collection. Suppose $G^+\subset g_1H\cup\cdots\cup g_kH$. Is it true that $H$ is of finite index in $G$?

Edit: as pointed out by Tom Goodwillie, the question is true without any assumption on $H$.

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You mean «$H<G$ the subgroup generated by $g$»? – Mariano Suárez-Alvarez Apr 7 '11 at 17:53
Mariano: Yes, this is what I mean. I've changed the statement, so there would be no misunderstanding. – Kate Juschenko Apr 7 '11 at 18:03
up vote 6 down vote accepted

Restating the question as I understand it:

Let $G$ be a group, $S\subset G$ a finite set of generators, $H$ a cyclic subgroup of $G$. Suppose that there is a finite collection of elements $g_j\in G$ such that every product of finitely many elements of $S$ belongs to one of the cosets $g_jH$. Is it true that $H$ has finite index in $G$?

The answer is yes, even without assuming $H$ cyclic. The group $G$ acts on the set $G/H$ of all cosets of $H$ in the usual fashion. The smallest subset of $G/H$ that contains the element $H$ and is closed under the action of each element of $S$ is finite. Therefore each element of $S$ maps this set to itself bijectively. Therefore so does the inverse of each element of $S$. Therefore so does each element of $G$. But the action of $G$ on $G/H$ is transitive, so it follows that this finite set is the whole of $G/H$, and $H$ has finite index.

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That was very fast! Thanks! – Kate Juschenko Apr 7 '11 at 18:20
You're welcome! – Tom Goodwillie Apr 7 '11 at 18:47
The finiteness of $S$ wasn't used, either, I realize. – Tom Goodwillie Apr 8 '11 at 21:17

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