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I have read the article http://www.math.upenn.edu/~kazdan/425S11/Drum-Gordon-Webb.pdf, , where Gordon and Webb describe in a simple a way the contruction of a pair of isospectral but non isometric plane domains.

At pag $52$ they mention the group $G$ used by Brooks and Buser in order to construct isospectral surfaces and they consider a particular set of generators of $G$ made by three elements, without specifying who is this group and these generators.

My question is this: is this $G$ the group $SL(3,2)$? Are the generators $\alpha=\left(\begin{array}{ccccccc} 0&0&1\\ 0&1&0\\ 1&0&0\\ \end{array}\right),$ $\beta=\left(\begin{array}{ccccccc} 1&0&0\\ 0&0&1\\ 0&1&0\\ \end{array}\right),$ $\gamma=\left(\begin{array}{ccccccc} 1&0&0\\ 0&1&0\\ 0&1&1\\ \end{array}\right) ?$

GGG
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    Everything that i stold about $G$, including the Cayley graph, supports this assumption (though I'd guess that their $\alpha$ is your $\gamma$) – Hagen von Eitzen Oct 11 '14 at 11:34
  • @ Hagen von Eitzen, thanks for your answer! Can you explain me why do you say that from the Cayley graphs it can be deduced that $G$ is very probably $SL(3,2)$? Why my $\alpha$ should be their $\gamma$? Thanks a lot for your help! – GGG Oct 11 '14 at 20:06
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    In http://www.math.ru.nl/~heckman/RS.Scriptie.pdf, section five explains a counterexample from Buser, Conway and Doyle (reference 4 at the end). This counterexample uses $PSL(\mathbb{F}_2^3)$, so the projective version of $SL(3,2)$. I don't currently have the time to see if this is the same counterexample, though. – HSN Oct 13 '14 at 11:52
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    Dumb question on my part, perhaps, but these aren't really Cayley graphs per se, are they? Just compact ways of encoding the permutations represented by $\alpha$, $\beta$ and $\gamma$... – Steven Stadnicki Mar 15 '15 at 16:40

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