Why is a Markoff number a third of a trace?

I fell down a rabbit hole this week and found myself thinking about Markoff numbers again.  I blogged about this before when Sarnak lectured here about them.  But I understood one minor point this week that I hadn’t understood then.  Or maybe I understood it then but I forgot.  Which is why I’m blogging this now, so I don’t forget again, or for the first time, as the case may be.

Remember from the last post:  a Markoff number is (1/3)Tr(A), where A is an element of SL_2(Z) obtained by a certain construction.  But why is this an integer?  Isn’t it a weird condition on a matrix to ask that its trace be a multiple of 3?  Where is this congruence coming from?

OK, here’s the idea.  The Markoff story has to do with triples of matrices (A,B,C) in SL_2(Z) with ABC = identity and which generate H, the commutator subgroup of SL_2(Z).  I claim that A, B, and C all have to have trace a multiple of 3!  Why?  Well, this is of course just a statement about triples (A,B,C) of matrices in SL_2(F_3).  But they actually can’t be arbitrary in SL_2(F_3); they lie in the commutator.  SL_2(F_3) is a double cover of A_4 so it has a map to Z/3Z, which is in fact the full abelianization; so the commutator subgroup has order 8 and in fact you can check it’s a quaternion group.  What’s more, if A is central, then A,B, and C = A^{-1}B^{-1} generate a group which is cyclic mod its center, so they can’t generate all of H.  We conclude that A,B, and C are all non-central elements of the quaternion group.  Thus they have exact order 4, and so their eigenvalues are +-i, so their trace is 0.

In other words:  any minimal generating set for the commutator subgroup of SL_2(Z) consists of two matrices whose traces are both multiples of 3.