The largest value of $k$ for $\Bbb{Z}^{k}$ to be embedded in $\mathcal{GL}(n,\Bbb{Z})$. Reading my course on  group theory, I asked my self the following question :

Suppose that $\Bbb{Z}^{k}$ can be embedded in $\mathcal{GL}(n,\Bbb{Z})$. What is the largest value of $k$?

 A: Derek was almost right: The precise answer is that the largest rank (call it $r$) of a free abelian subgroup of $SL(n,Z)$  equals 
$$
d=\left[\frac{n^2}{4} \right] 
$$
where the bracket is the floor function. The proof is a bit heavy. To see that $r\ge d$ you construct a free abelian subgroup $H$ of rank $d$ as the subgroup of matrices of the form 
$$
I + (a_{ij})
$$
where $I$ is the identity matrix and $a_{ij}\in Z$ is zero provided $i(j+1)=0$ modulo $2$. This means that the matrix $A=(a_{ij})$ has the form
$$
\left[\begin{array}{cccc}
0 & * & 0 & * ...\\
0 & 0 & 0 & ...\\
0 & * & 0 & * ...\\
& & \ldots  & &
\end{array}
\right]
$$
Hence, each matrix $A=(a_{ij})$ has exactly $d$ entries which are, potentially, nonzero. I leave you to check that $H$ is indeed a subgroup of $SL(n,Z)$ and is isomorphic to $Z^d$. 
(The key is that $(I+A)(I+A')= I+ A + A'$ for all matrices $A, A'$ of the above type.) 
To get the upper bound, you have to use some heavy machinery (or not so heavy if you saw this staff before): If $H$ is a free abelian subgroup of $SL(n,Z)$ of rank $k$, you take the Zariski closure $\bar{H}$ of $H$ in $SL(n,R)$; the result is isomorphic to the additive group $R^k$. Then you take the Lie algebra of $\bar{H}$; it is a rank $k$ commutative subalgebra $a$ of the Lie algebra $g$ of traceless $n\times n$ real matrices. Then you do a bit of googling and discover this conversation, from which you learn that the maximal dimension of $a\subset g$ is exactly $d$ (it seems that the result was first proven by I.Schur). Therefore, $r\le d$. 
Combining this, you obtain that $r=d$. Lastly, since the group $SL(n,Z)$ has finite index in $GL(n,Z)$, the ranks of their maximal free abelian subgroups are the same.  
