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Nov 15, 2016 at 11:35 comment added François Labourie Indeed, there was a typo here and the correct (and needed) conclusion is the one stated by Dave.
Mar 28, 2015 at 6:26 vote accept Cusp
Mar 27, 2015 at 18:47 comment added Dave Witte Morris The key point in my answer is that I moved the assertion "$h \in \Gamma_n$" from the conclusion to the hypotheses.
Mar 27, 2015 at 16:14 comment added Dave Witte Morris To simplify, suppose the image of $H_m$ in $\langle\eta\rangle \backslash \Gamma /\langle\gamma\rangle$ has only one double coset $\langle\eta\rangle h \langle\gamma\rangle$ that is not in $\langle\eta\rangle \cdot \langle\gamma\rangle$. By double coset separability, there exists $\Gamma_m'$, such that $h \notin \langle\eta\rangle \Gamma_m' \langle\gamma\rangle$. Since $\langle\eta\rangle h \langle\gamma\rangle$ is the only one double coset that is not in $\langle\eta\rangle \cdot \langle\gamma\rangle$, this implies $H_m \cap \Gamma_m' \subseteq \langle\eta\rangle \cdot \langle\gamma\rangle$.
Mar 27, 2015 at 16:12 comment added Dave Witte Morris You are absolutely right that double coset separability only tells us that something is not in $\Gamma_n$. (So the original conclusion cannot possibly be correct.) However, telling us something is not in $\Gamma_n$ is the same as putting a restriction on the things that are in $\Gamma_n$.
Mar 27, 2015 at 8:47 comment added Cusp Why double coset separability implies that there is a finite-index normal subgroup $\Gamma_m'$, such that $H_m \cap \Gamma_m' \subset \langle \eta_m \rangle \cdot \langle \gamma_m \rangle$? From my point of view double coset separability tells you what is not in $\Gamma_n$. How does it implies something is in $\Gamma_n.$
Mar 27, 2015 at 8:35 history edited Dave Witte Morris CC BY-SA 3.0
corrected an oversight -- need to consider all possible eta and gamma
Mar 27, 2015 at 7:30 history answered Dave Witte Morris CC BY-SA 3.0