What I am going to ask is probably simple and maybe trivial. But I want to be sure that I am not missing any point. Let ${\bf G}\subseteq \mathrm{GL}\_n(\mathbb{C})$ be a simple and simply connected group. Then this group is defined over $\mathbb{Z}$. Hence I can talk about ${\bf G}(\mathbb{F}\_q)$ where the characteristic of the finite field $\mathbb{F}\_q$ is $p$. 


1. Is it true that when $p\geq 3$, the finite group ${\bf G}(\mathbb{F}\_q)$ is perfect? Of course since $\bf G$ is simple then $\bf G=\bf G'$ but does this imply that the commutator subgroup of ${\bf G}(\mathbb{F}\_q)$ is the same as ${\bf G}(\mathbb{F}\_q)$.
2. Moreover is it true that ${\bf G}(\mathbb{F}\_q)/(Z({\bf G}(\mathbb{F}\_q)))$ is a simple group?


Thanks



Simple groups of Lie type
I'll just make my comment an answer so as to close this question. By Theorem 24.17 of the book "Linear Algebraic Groups and Finite Groups of Lie Type" by Malle and Testerman we have the answer is yes unless $\mathbf{G}(\mathbb{F}\_q)$ is one of the groups


$\mathrm{SL}\_2(\mathbb{F}\_2)$, $\mathrm{SL}\_2(\mathbb{F}\_3)$, $\mathrm{SU}\_3(\mathbb{F}\_2)$, $\mathrm{Sp}\_4(\mathbb{F}\_2)$, $\mathrm{G}\_2(\mathbb{F}\_2)$, ${}^2\mathrm{B}\_2(\mathbb{F}\_2)$, ${}^2\mathrm{G}\_2(\mathbb{F}\_3)$, ${}^2\mathrm{F}\_4(\mathbb{F}\_2)$.


These groups are genuine exceptions as is explained in Remark 24.18.


