Reductive groups over finite fields Let $\ell\ge 5$ be a prime number. Let $G/\mathbb{F}_\ell$ be a (smooth, connected) reductive algebraic group. Let $G(\mathbb{F}_\ell)^+$ be the normal subgroup of $G(\mathbb{F}_\ell)$ generated by its $\ell$-Sylow subgroups. Is it true that $G(\mathbb{F}_\ell)/G(\mathbb{F}_\ell)^+$ is abelian? 
Comments:
(1) An answer ``for sufficiently large $\ell$'' would suffice for my purpose.
(2) I seem to understand that the group $G(\mathbb{F}_\ell)/G(\mathbb{F}_\ell)^+$ is solvable.
 A: This is a complement to Jim Humphreys' answer, describing the problem from the perspective of a general ground field $k$ (after some preliminaries over finite fields). Let $G$ be a connected reductive group over any field $k$, with connected semisimple derived group $G'$, and let $\widetilde{G}' \rightarrow G'$ be the simply connected central cover (over $k$!). 
If $k$ is finite of characteristic $\ell$
then any $\ell$-subgroup of $G(k)$ may be regarded as a unipotent constant finite $k$-subgroup of $G$, so by a theorem of Borel and Tits (that is addressed in the MO question "homomorphisms of reductive groups") such subgroups lie in connected unipotent smooth closed $k$-subgroups of $G$. These are $k$-split since $k$ is perfect. By the Borel-Tits structure theory (over any field), split connected unipotent smooth closed $k$-subgroups lie in unipotent radicals of minimal parabolic $k$-subgroups, so for finite $k$ their groups of $k$-points are exactly the Sylow $\ell$-subgroups of $G(k)$.  
In other words, what you are calling $G(k)^+$ has a definition (due to Tits) that makes sense over any field: it is the (visibly normal!) subgroup of $G(k)$ generated by $U(k)$'s as $U$ varies through the unipotent radicals of the minimal parabolic $k$-subgroups of $G$. This is alluded to in Jim Humphreys' answer.  So now it makes sense to ask about the commutativity of $G(k)/G(k)^+$ for any field $k$. 
Since $G/G'$ is a torus, obviously 
$G(k)^+ = G'(k)^+$. Since $\widetilde{G}' \rightarrow G'$ is a central quotient map between connected semisimple groups, it restricts to isomorphisms between unipotent radicals of minimal parabolic $k$-subgroups (by the Borel--Tits relative structure theory), so $\widetilde{G}'(k)^+ \rightarrow G'(k)^+$ is surjective.  Thus,
$$G(k)/G(k)^+ = {\rm{coker}}(\widetilde{G}'(k)^+ \rightarrow G(k)).$$
In other words, does the commutator subgroup of $G(k)$ lie in the image of $\widetilde{G}'(k)^+$?
Now comes a step that uses algebraic geometry quotients in an essential manner: I claim that the commutator morphism of varieties $c:G \times G \rightarrow G$ factors through the natural $k$-homomorphism $f:\widetilde{G}' \rightarrow G$.  The crux of the matter is that $f$ induces an isomorphism between maximal central quotients; i.e., it maps $\widetilde{G}'/Z_{\widetilde{G}'}$ isomorphically onto $G/Z_G$. Consequently, for any $g_1, g_2 \in G(A)$ for a $k$-algebra $A$, fppf-locally on Spec($A$) the images $\overline{g}_i \in (G/Z_G)(A)$ lift to points $\widetilde{g}_i$ of $\widetilde{G}'$ (valued in an fppf $A$-algebra $A'$), with the lift ambiguous up to translation by $Z_{\widetilde{G}'}$. Such  central translations wipe out when forming the commutator $(\widetilde{g}_1, \widetilde{g}_2) \in \widetilde{G}'$, so by descent theory we have defined a $k$-scheme morphism $G \times G \rightarrow \widetilde{G}'$ that does the job.
This proves the claim.  (Of course, this claim is known by more classical means, but I think that the viewpoint of fppf group sheaves provides the only proof which proceeds "as if" these were ordinary groups without getting hung up on special contortions due to inseparable isogenies.) 
OK, the algebraic geometry has done its work, and we conclude that $G(k)/G(k)^+$ is commutative provided that $\widetilde{G}'(k)^+ = \widetilde{G}'(k)$. In other words, the answer (now over a general field) is affirmative provided that the analogous quotient for $\widetilde{G}'$ is trivial. In other words, we are reduced to a refinement of the original question (over the same ground field we started with): if $G$ is simply connected semisimple then is $G(k)^+ = G(k)$?
Well, we can assume $G \ne 1$, so $G = \prod {\rm{R}}_{k_i/k}(G_i)$ (Weil restrictions) for some finite separable extension fields $k_i/k$ and absolutely simple and simply connected semisimple $k_i$-groups $G_i$. The bijection between parabolic $k$-subgroups of $G$ and collections of such $P_i \subset G_i$ (via Weil restriction) thereby identifies $G(k)^+$ with $\prod G_i(k_i)^+$, so at the cost of working over each $k_i$ separately we may arrange moreover that $G$ is absolutely simple.  Moreover, in the case of finite $k$ we know that $G$ is isotropic (since there's always  Borel subgroup defined over the ground field).
The upshot is that we are reduced to the following question (as the cost of replacing
the original ground field with finite separable extensions, so a good reason not to have restricted only to prime finite fields at the start): if $G$ is a connected semisimple $k$-group that is simply connected, absolutely simple, and $k$-isotropic then does $G(k)^+ = G(k)$?  This question is called the "Kneser-Tits Conjecture", and it has an interesting history (and its own Wikipedia page). It can fail over some fields, and was recently solved over global fields in the final thorniest cases in the affirmative. The special case of finite fields was settled a long time ago by Steinberg, who proved it more generally in the quasi-split case over any field (which for finite fields is the general case).
Sorry for rambling so much about this, but I think it is important to recognize that irrespective of the theory of BN-pairs, the true underlying problem is a question of an equality in the absolutely simple simply connected case (over a possibly bigger finite field). Of course, to attack this refined problem over finite fields, or more generally in the quasi-split case over any field, BN-pairs are the tool of choice (and the answer is always affirmative). 
A: The quotient you describe is certainly abelian thanks to the standard BN-pair structure, though your conditions are overly restrictive.  I'm not sure why you use the letter $\ell$ for the defining prime characteristic, or why you work only over the prime field, or why you avoid some primes.    (Some discussion of motivation would help.)    Usually one deals with connected reductive groups over a finite field $\mathbb{F}_q$ with $q$ a power of a prime $p$.   Often the letter $\ell$ is used for a different prime dividing the finite group order, when there is discussion of representations other than in the defining characteristic.    
The essential case is that of a (connected) semisimple algebraic group defined over a finite field $k$, since a (connected) reductive group is the almost-direct product of a (commutative) torus and such a semisimple group.   Then the Borel-Tits structure theory, over an arbitrary field of definition, specializes nicely to a finite field and yields a Bruhat decomposition (encoded axiomatically in a BN-pair).    From this it's clear that your (normal) subgroup $G(k)^+$ is generated by rational points of unipotent radicals of Borel subgroups defined over the finite field.  Moreover, the only elements of $G(k)$ which might not lie in $G(k)^+$ must lie in the torus part, which is commutative.    
These ideas were developed in more generality by Tits in his article: Algebraic and abstract simple groups, Ann. of Math. 80 (1964), which is relied on in numerous papers and some books in the study of finite simple groups.     Tits was responsible for the $G^+$ notation in greater generality, leading to further study of this subgroup over arbitrary fields especially when $G$ is simply connected.     
A: Let B=TU be a Borus of G defined over Fq. Let H be the image of T(Fq) in G(Fq)/G(Fq)+. If H has index k, then the preimage of H in G(Fq) has index k. The preimage of H contains the large cell U-BU, which is dense in G. At least for a fixed type of G and q>>0, this immediately implies that k=1 (which of course implies the required abelianility). In general, a finer analysis of the formula for the number of points in a reductive group over a finite field should get finer control over for which q and G this argument works.
Actually (Alternatively, using Bruhat and idea of looking at image of T(Fq) it suffices to show that every Weyl group element has a representative in the subgroup generated by unipotent elements. The Weyl group is generated by simple reflections, so this question is reduced to a computation in rank one, where it suffices to consider the simply connected groups SL2 and SU3.
