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](http://arxiv.org/abs/math/0210033) 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?



Product of conjugate matrices in $\mathrm{SL}(2, \mathbb{Z})$
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 $12 \mid n$.


