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Jun 8, 2010 at 13:59 comment added Peter Samuelson This is a very clean answer (at least, modulo the proof that $SL_2(\mathbb{Z}) \cong Z_4 \ast_{Z_2} Z_6$, which I knew already). Also, the fact that you mention feels like it should be true, and it's nice to know that it is.
Jun 8, 2010 at 13:55 vote accept Peter Samuelson
Jun 7, 2010 at 0:11 comment added Autumn Kent That's right. It follows, in particular, from the more general Kurosh Subgroup Theorem. See en.wikipedia.org/wiki/Kurosh_subgroup_theorem
Jun 6, 2010 at 23:40 history answered Allen Knutson CC BY-SA 2.5