Let $G$ be a finite group, and suppose that a set of proper subgroups $H_1,\dotsc,H_n$ satisfy $G=\bigcup_{g\in G}\bigcup_{i=1}^nH_i^g$, where $H_i^g$ is the conjugate of $H_i$ by $g$. In this case, we say that $G$ is conjugate-covered by the $H_i$. (Note, sometimes this is called a normal cover of $G$, but sometimes this seems to mean something else.)
What can be said about the $H_i$ when they conjugate-cover $G$, besides the obvious fact that the size of the above union is equal to $|G|$?
There are many results on the minimal size of a conjugate-cover of a group, or when a conjugate-cover with some special properties (nilpotent, Sylow, …) exists. However, I want the reverse — start with a group $G$ of interest, and find minimal (or at least fairly general) sufficient conditions on a set of subgroups $H_i$ so that they conjugate-cover $G$. These sorts of results seem rather more sparse.
I know of a few that came up in this question concerning the case where the $H_i$ are all normal in $G$, so the conjugation does nothing. The strongest of this nature seem to be:
Theorem. A group has a nontrivial finite covering by normal subgroups if and only if it has a quotient isomorphic to an elementary Abelian $p$-group of rank two for some prime $p$.
Corollary. Let $G=\bigcup_{i=1}^nN_i$ where $N_1,\dotsc,N_n$ form an irredundant covering of $G$ by proper normal subgroups. Then $G/D$ with $D=\bigcap_{i=1}^nN_i$ is finite and solvable.
Note, a covering is called irredundant if no proper subset is also a covering.
The reference is
Brodie, M. A., Chamberlain, R. F., & Kappe, L.-C. (1988). Finite Coverings by Normal Subgroups. Proceedings of the American Mathematical Society, 104(3), 239–258.
There is also a nice paper on partitions of groups (coverings where the pairwise intersections are trivial) which contains the following, that also constitute a proof of the "pretty" fact mentioned at the above link:
Theorem. Let $G$ be a finite group partitioned by $\Pi$. Assume that $HK=KH$ for all $H,K\in\Pi$. Then $G$ is an elementary Abelian $p$-group.
Theorem. Let $G$ be finite and partitioned by $\Pi$. Suppose $A\in\Pi$ and $AH=HA$ for all $H\in\Pi$. Then $A\trianglelefteq G$.
Reference for these:
Isaacs, I. M. (1973). Equally Partitioned Groups. Pacific Journal of Mathematics, 49(1), 109–116.
Finally, a seemingly-lesser-known result of Artin rephrases this in terms of group characters:
Artin's Theorem on Induced Characters. Let $G$ be a finite group. $G$ is conjugate-covered by $H_1,…,H_n$ if and only if every finite-dimensional, complex character of $G$ is a $\mathbb{Q}$-linear combination of induced characters from the $H_i$.
See, for example, Kramár - Artin's and Brauer's theorems on induced characters.
This seems a powerful result, but I'm too inexperienced in character theory to know what to do with it.
My motivation for this problem is in studying intersective polynomials, which are polynomials over $\mathbb{Z}$ which have roots in $\mathbb{Z}/n\mathbb{Z}$ for all positive $n$ (of course, the really interesting ones are those without roots in $\mathbb{Q}$).
Berend & Bilu showed (citation to follow) that the Galois group of such a polynomial must be conjugate-covered by certain subgroups, one for each irreducible factor of the polynomial (the subgroups are those fixing a root of one irreducible factor). In particular, such a polynomial must be reducible of degree $\geq 5$.
Berend, D., & Bilu, Y. (1996). Polynomials with Roots Modulo Every Integer. Proceedings of the American Mathematical Society, 124(6), 1663–1671.
In low ($5$ and $6$) degrees it's not hard to show that most Galois groups of reducible polynomials cannot be covered in this way, but the problem balloons for higher degrees, so knowing more about the constraints placed on $G$ and the $H_i$ would be useful.
