Union of conjugates of a subgroup Let $G$ be a finite group, $H \leq G$ a proper subgroup. It is well known that the union of the conjugates of $H$ does not cover $G$. I would like to know of more precise results (even in special cases) saying that the union of conjugates misses some structure.
For example, if $G$ is a Frobenius group, the complement of the union of conjugates is a subgroup (when $1$ is added). Is there any generalization of it? Maybe one could hope to find a submonoid (in the complement) in a more general case?
I would like to be able to deduce that this union misses something more specific than just some element
It is possible that there is something we can say if $G$ is a $p$-group.
 A: To avoid examples like subgroups of cyclic groups, we could assume that $H$ is core-free in $G$. Then the question is equivalent to asking about fixed point free elements of transitive permutation groups.
In a transitive $p$-group, a minimal normal subgroup has order $p$ and must be fixed-point-free (since the stabilizer is core-free), so the complement (plus identity) contains a subgroup of order $p$.
In general, there need be no fixed-point-free elements of prime order. For example, TransitiveGroup(12,47) (in the GAP/Magma database) is a group of order $72$ with stabilizer dihedral of order $6$. There are $54$ fixed-point-free elements, all of order $4$, so the complement plus identity contains no submonoid.
A: This does not really answer the question, but perhaps it is worth pointing out that if $H$ is proper in $G$, then the number of elements of $G$ missed by the union of the conjugates of $H$ is at least the order of $H$. This is elementary, but a deeper result, proved using an appeal to the simple group classification is that the set of missed elements always contains an element of prime-power order
A: I'm not sure whether it is really relevant, but there is a generalization by Wielandt of Frobenius's theorem which may be interpreted as follows: Let $G$ be a transitive permutation group on a  set $\Omega$ , and let $G_{\alpha}$ be a point stabilizer. Suppose that $G_{\alpha} \neq \langle G_{\alpha} \cap G_{\beta} : \beta \neq \alpha \rangle.$ Then there is a proper normal subgroup of $G$ which is still transitive on $\Omega.$
This is a permutation group theoretic consequence  of Wielandt's generalization of Frobenius's theorem : if $H$ is a subgroup of a finite group $G$ and there is $K \lhd H$ with $H \cap H^{g} \leq K$ for all $g \in G \backslash H,$ then there is $L \lhd G$ with $G = HL$ and $L \cap H = K.$
