Let me introduce the context:
A few week ago I have made the following contest as a "homework" : ENS contest (France) 2006 which is essentially about $SL_n(\mathbb{Z})$ group, finite subgroup of $SL_n(\mathbb{Z})$ and as a final question : Any surjective morphism group $SL_n(\mathbb{Z}) \rightarrow SL_n (\mathbb{Z})$ is bijective.
Yesterday I had proved that $S_n$ is isomorphic to a subgroup of $\mathrm{GL}_{n-1}(\mathbb{Z})$.
Which is mainly a consequence of the mentioned contest.
So, here is my question :
Assuming that $S_n$ has an element of order $m\in \mathbb{N}$. Does the following group $\mathrm{GL}_{n-2}(\mathbb{Z})$ has an element of order $m$ ?
For $n=\{1,2,3\}$ one can check the answers is no.
Unfortunately, I had no idea to disprove (perhaps the result is true for $n \geq 4$) the claim for higher values of $n$.
EDIT: It seems to be true for $n\ge 4$ and even. (Look at Greg's Martin answer).
One should be able to enumerate all possible finite orders of elements of GL$_k(\Bbb Z)$. The characteristic polynomial would have to be a product of cyclotomic polynomials $\Phi_{q_i}$, and the order of the matrix would be the lcm of the $q_i$. Since the degree of the characteristic polynomial is (bounded by) $k$, this yields a finite list of possible orders. For $k=2$, the possible orders are $\{1,2,3,4,6\}$, so I think the answer is yes for $n=4$. But the answer is no for $n=5$, since no $3\times3$ matrix can have order $5$. (Indeed, the answer is no for any prime $n$.)