I just came to this conjecture (proved by M.Raynaud and D.Harbater in 1994) last weekend, in Fresnel and v.d.Put's book Rigid Geometry and Its Applications. It claims that all quasi $p$group $G$ could be characterized as certain Galois group of a Galois covering $Y $to $\mathbf{P}_1$, only ramified at infinity (over algebraic closed field with positive character). Since many grandmasters have researched this conjecture, what is the importance of Abhyankar's conjecture ? Well, of course it relates to the inverse Galois theory...

4$\begingroup$ The statement you give is Raynaud's contribution. Harbater's theorem is for covering of arbitrary algebraic curves — it says that a group $G$ is the Galois group of a covering of a curve (genus $g$, minus $r$ points) over an algebraically closed field of characteristic $p$ iff its quotient $G/p(G)$ by the subgroup $p(G)$ generated by $p$subgroups is such a group. By Grothendieck the latter property means that it is a quotient of the fundamental group of a compact Riemann surface of genus $g$ deprived of $r$ points. $\endgroup$– ACLJul 6, 2015 at 12:46

$\begingroup$ Yes, I think what you said is the strong version of Abhyankar's conjecture. $\endgroup$– YUAN ZhiriJul 6, 2015 at 12:57
2 Answers
For a bit of context, recall that the affine line over an algebraically closed field of characteristic zero is simply connected in the sense that it has no nontrivial étale covers. (You can see this using RiemannHurwitz or by reducing to the corresponding topological statement over $\mathbb{C}$.) In characteristic $p>0$ this is false! For example, $\mathbb{A}^1_k$ has ArtinSchreier covers. This raises the question of what $\pi_1(\mathbb{A}^1_k)$ looks like. By construction, this is an inverse limit of finite groups. The weaker question of what these finite groups are is completely answered by the (solution to) Abhyankar's conjecture: these are precisely the quasi$p$ groups (groups generated by the Sylow $p$subgroups). I think that this is a pretty remarkable outcome.

$\begingroup$ @YuanZhiri Also note that what RiemannHurwitz gets you is that $\mathbb A^1$ has no nontrivial stale covers that have Galois group of order prime to $p$, which implies that every finite cover must have a Galois group generated by $p$subgroups. So the solution shows that RiemannHurwitz is the only obstruction. $\endgroup$ Jul 6, 2015 at 14:05

1$\begingroup$ @WillSawin, you have an (autocorrect?) error in the type of covers described in your comment. $\endgroup$– KConradJul 6, 2015 at 14:16

$\begingroup$ @WillSawin Excuse me I am not familiar with etale coverings. So for etale covering f:Y to A^1, the RiemannHurwitz implies that \chi(Y)=\chi(A^1)*deg(f) for unramifiedness. Then what should I do after this ? (It seems that the Euler character of Y contains a power of p.) Where could I find these details ? $\endgroup$ Jul 6, 2015 at 15:37
This is not an answer to the question, "What is the deeper meaning of Abhyankar's conjecture?" It is an answer to the question, "What is one application of Abhyankar's conjecture?" Over an algebraically closed field $k$ of characteristic $p$, every $k$action of a finite $p$group on every proper, separably rationally connected $k$variety (e.g., every rational $k$variety) has a $k$point fixed by the entire group. In particular, every finite $p$group in every semisimple group of adjoint type is contained in a Borel subgroup (maybe there is a direct proof of this, but I do not know it).
The proof, following an argument introduced by Kollár and Debarre, combines Raynaud's solution to Abhyankar's conjecture with the theorem of de Jong and myself (the positive characteristic generalization of the theorem of Graber, Harris and myself). Please confer the following answer of ChambertLoir as well as my comment, Are rational varieties simply connected?.
Edit. There are two remarks. First, once the $p$group is contained in a Borel subgroup, automatically it is contained in the unipotent radical of that Borel subgroup, since the multiplicative quotient has trivial $p$torsion group (the $p$torsion group scheme structure, of course, is nontrivial). Second, this result completely fails in characteristic prime to $p$. For every integer $n$ that is divisible by $p$, for every algebraically closed field $k$ of characteristic prime to $p$, there is a copy of $\mathbb{Z}/p\mathbb{Z}\times \mathbb{Z}/p\mathbb{Z}$ in $\textbf{PGL}_{n,k}$ that is contained in no Borel subgroup. There is a lift of this group to a $p$group of order $p^3$ in $\textbf{SL}_{n,k}$.
Second edit. Poonen explained to me a counting argument for the $p$group above: first reduce to the case of finite fields of characteristic $p$, and then observe that the number of rational points of the flag variety of $G$ is congruent to $1$ modulo $p$. Thus there must be an orbit of size $1$. However, the argument above also applies to groups of the form $Q\times \mathbb{Z}/\ell\mathbb{Z}$, where $P$ is a finite $p$group and $\ell$ is an integer prime to $p$. The counting argument does not imply this case, but Abhyankar's conjecture does. Also, for flag varieties, Lang and Steinberg proved the existence of rational points in the 1960's.