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.

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    $\begingroup$ In a finite $p$-group maximal subgroups are normal and $H$ is contained in a maximal subgroup. So probably the result will differ between nilpotent groups and other groups. $\endgroup$ – j.p. Jul 8 '14 at 10:42
  • $\begingroup$ I think for abelian groups there's no hope since the minimal submonoid containing the complement is the full group. $\endgroup$ – Cristos A. Ruiz Jul 8 '14 at 10:46
  • $\begingroup$ The question does not make sense when $G$ is abelian (or $H$ normal) since then the union is $H$. $\endgroup$ – Lior Bary-Soroker Jul 8 '14 at 10:57

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.

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    $\begingroup$ Michael Guidici's work on fixed-point-free elements is relevant here. Note that these elements are also sometimes called derangements, and a transitive permutation group is called elusive if it contains no derangements of prime order. Michael has some very strong results towards a classification of elusive permutation groups. $\endgroup$ – Nick Gill Jul 8 '14 at 15:10
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    $\begingroup$ ... For instance, one particularly nice result of Michael's: The only almost simple elusive groups are $M_{11}$ and $M_{10}=A_6.2$ acting on 12 points. A nice summary is here: staffhome.ecm.uwa.edu.au/~00033321/anztalk.pdf $\endgroup$ – Nick Gill Jul 8 '14 at 15:11

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


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 $L \cap H = K.$


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