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I've read that if $M_1, \dots, M_n$ are matrices in $\mathrm{SL}(2, \mathbb{Z})$ whose product is the identity, and each is conjugate to the shear

$$ \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}, $$

then $n$ is a multiple of $12$. This appears, for example, at the bottom of p25 of this paper of Symington, with references to two books on four-manifolds. Is there a simple algebraic proof of this fact, or intuition for why the number 12 should be involved?

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    $\begingroup$ ${\rm SL}(2,\mathbb{Z})$ has a free normal subgroup of index 12 such that the factor group is cyclic. $\endgroup$ – Geoff Robinson May 8 '16 at 9:48
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See Keith Conrad's notes http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf, particularly Example 2.5. Let us write (as Conrad does) $S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Then $S$ has order $4$ and $ST = \begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}$ has order $6$. In his Example 2.5, Conrad displays a character $\chi : SL_2(\mathbb{Z}) \to \mathbb{C}^*$ taking $S$ to a primitive $4$th root of unity and $ST$ to a primitive $6$th root of unity. So $\chi(T) = \chi(S)^{-1} \chi(ST)$ is a primitive $12$th root of unity; in fact Conrad's example has $\chi(T) = e^{2\pi i/12}$. And $\chi(M)$ has this same value for any matrix $M$ conjugate to $T$.

At this point perhaps it is obvious, but to be clear: if $M_1,\dotsc,M_n$ are each conjugate to $T$ and $M_1 \dotsm M_n$ is the identity, then $1 = \chi(\prod M_i) = \prod \chi(M_i) = \chi(T)^n = e^{2 \pi i \, n/12}$, so $12 \mid n$.

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To expand my comment, and combine it with some points from Zach Teitler's answer, but more in a generators and relations framework: ${\rm SL}(2,\mathbb{Z})$ is well-known to be isomorphic to the group $G = \langle s,t: s^{4} = (st)^{6} = 1, s^{2} = (st)^{3} \rangle $. But notice that the cyclic group $C = \langle c \rangle $ of order $12$ satisfies these relations ( with an element of order $4$ in the role of $s$ and an element of order $12$ in the role of $t$).

Hence there is a homomorphism $\phi: G \to C$ with $\phi(s) = c^{-3}, \phi(t) = c.$ Since $G/C$ is Abelian, $\phi$ contains the derived group $[G,G]$ in its kernel so all $G$-conjugates of $t$ have the same image $c$ under $\phi$.

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  • $\begingroup$ This is just the sort of thing I was looking for; I would have accepted both answers if I could :) $\endgroup$ – Jez May 8 '16 at 16:19
  • $\begingroup$ Some of the above relations are redundant: it would be better to write $G = \langle s,t : s^{4} = 1, s^{2} = (st)^{3} \rangle$. $\endgroup$ – Geoff Robinson Jun 7 '16 at 8:01
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Since $\mathbb{Z}/12 \cong \mathbb{Z}/3 \times \mathbb{Z}/4$, it is enough to see how to build surjections $SL(2,\mathbb{Z}) \to \mathbb{Z}/3$ and $SL(2,\mathbb{Z}) \to \mathbb{Z}/4$. The former is easy: $SL(2,\mathbb{Z})$ acts on $\mathbb{P}^1(\mathbb{Z}/3)$. This is a permutation action on a four element set, and a bit of computation shows that the image is the alternating group $A_4$. We can compose $$SL(2, \mathbb{Z}) \to A_4 \to \mathbb{Z}/3.$$

I can't find an equally nice description of the map $SL(2, \mathbb{Z}) \to \mathbb{Z}/4$. It factors through $SL(2, \mathbb{Z}/4)$, but I can't see how to show that there is a map $SL(2, \mathbb{Z}/4) \to \mathbb{Z}/4$ in any way more elegant than writing it down.

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  • $\begingroup$ That's a very nice description of the map to $\mathbb{Z}/3$! $\endgroup$ – Jez May 12 '16 at 9:52

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