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Mar 18, 2010 at 22:04 comment added Victor Miller First you should work in $PSL_2(\mathbb{Z})$ just to get rid of the annoying $\pm 1$. As you stated the result is that $PSL$ is the free product of the images of $S_4$ and $S_6$. This means that every element has a unique representation as a reduced word in them: i.e a word of the form $S_4^{a_1} S_6^{b_1} \dots S_4^{a_n} S_6^{b_n}$ where $a_i = 1$ and $b_i \in \{1,2\}$ and we can optionally omit the leading and/or trailing factors. From this it's easy to solve any word problem in your $a$ and $b$ (almost as easy as in a free group)
Mar 18, 2010 at 21:44 vote accept Peter Samuelson
Mar 18, 2010 at 21:07 answer added Andy Putman timeline score: 6
Mar 18, 2010 at 7:42 history edited Peter Samuelson CC BY-SA 2.5
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Mar 18, 2010 at 7:34 comment added Douglas Zare You have to use an extra \ to get the matrices to display correctly. Similarly, to display { } use \\{ \\}.
Mar 18, 2010 at 7:31 history edited Douglas Zare CC BY-SA 2.5
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Mar 18, 2010 at 7:26 answer added Robin Chapman timeline score: 11
Mar 18, 2010 at 7:25 answer added Jack Schmidt timeline score: 6
Mar 18, 2010 at 7:13 history asked Peter Samuelson CC BY-SA 2.5