Frattini and Derived Subgroup of $SL(n,\mathbb{Z})$

It is known that the derived subgroup of $SL(2,\mathbb{Z})$ is subgroup of index 12 in the group.

1. What is known about derived subgroup of $SL(n,\mathbb{Z})$, for $n\geq 3$? (is it finitely generated?, is its index is finite in group? etc.)

2. What is the Frattini subgroup of $SL(n, \mathbb{Z})$, for $n\geq 2$?

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I believe that ${\rm SL}(n,\mathbb Z)$ is perfect for $n \ge 3$. You can get a transvection as a commutator:
$\left[ \left( \begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) , \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{array} \right) \right] =\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array} \right).$
and similarly you can get any matrix which is the identity with a single $1$ elsewhere. It is not hard to show that ${\rm SL}(n,\mathbb Z)$ is generated by matrices of this form.
I am sure the Frattini subgroup must be trivial for all $n \ge 2$. The kernels $K_p$ of the natural maps onto ${\rm SL}(n,p)$ for primes $p$ all equal to intersections of maximal subgroups, and the intersection of all of the $K_p$ is clearly trivial.
Think of the above commutator relation as $[(2,1),-(3,2)] = (3,1)$. Then for any distinct $1 \le i,j,k \le n$, we have $[(j,i),-(k,j)]=(k,i)$. –  Derek Holt Mar 7 '13 at 11:22