Sylow's theorem with operators

Statement
Suppose $$G$$ is a finite group and $$H$$ is a coprime automorphism group of $$G$$: $$H$$ is a subgroup of $$\operatorname{Aut}(G)$$ such that the orders of $$G$$ and $$H$$ are relatively prime. A subgroup of $$G$$ is termed $$H$$-invariant if it equals its image under any element of $$H$$.

Suppose further that either $$G$$ or $$H$$ is solvable.

Let $$p$$ be any prime.


 * 1) Existence (E): The set of $$H$$-invariant $$p$$-Sylow subgroups of $$G$$ is nonempty.
 * 2) Conjugacy (C): Any two $$H$$-invariant $$p$$-Sylow subgroups of $$G$$ are conjugate by an element in $$C_G(H)$$.
 * 3) Domination (D): Any $$H$$-invariant $$p$$-subgroup of $$G$$ is contained in a $$H$$-invariant $$p$$-Sylow subgroup of $$G$$.

Note that since given two groups of coprime order, one of them is solvable, the assumption that either $$G$$ or $$H$$ is solvable is redundant.

Facts used

 * 1) uses::Sylow subgroups exist
 * 2) uses::Frattini's argument
 * 3) uses::Normal Hall implies permutably complemented
 * 4) uses::Hall retract implies order-conjugate: This states that any two complements to a normal Hall subgroup are conjugate. The proof of this is what requires the assumption that either the normal Hall subgroup or the quotient group is solvable.
 * 5) uses::Normal Hall satisfies transfer condition

Proof of existence
Given: A finite group $$G$$, a subgroup $$H$$ of $$\operatorname{Aut}(G)$$ such that $$G$$ and $$H$$ have relatively prime orders. A prime $$p$$.

To prove: There exists a $$H$$-invariant $$p$$-Sylow subgroup of $$G$$.

Proof: Let $$K$$ be the semidirect product of $$G$$ with $$H$$. $$G$$ is a normal Hall subgroup of $$K$$, and $$G/K \cong H$$.


 * 1) By fact (1), $$G$$ has a $$p$$-Sylow subgroup, say $$P$$.
 * 2) Let $$N = N_K(P)$$. Then, $$K = GN$$: This follows from fact (2), and the fact that $$G$$ is normal in $$K$$.
 * 3) There exists a permutable complement $$L$$ to $$G \cap N$$ in $$N$$: Since $$G$$ is normal Hall in $$K$$, $$G \cap N$$ is normal Hall in $$N$$. Fact (3) thus applies.
 * 4) $$L$$ is a permutable complement to $$G$$ in $$K$$: Since $$GN = K$$, we get $$G((G \cap N)L) = K$$. Since $$G \cap N \le G$$, this yields $$GL = K$$. Further, $$L \le N$$, so $$G \cap L \ge G \cap L \cap N = (G \cap N) \cap L$$, which is trivial from the previous step.
 * 5) $$L$$ is conjugate to $$H$$ by an element of $$K$$. Since $$GH =K$$, $$L$$ and $$H$$ are conjugate via an element of $$G$$: This follows from fact (4), and the previous step, which shows that $$L$$ is also a permutable complement to $$G$$ in $$K$$.
 * 6) Let $$g$$ be the element of $$G$$ conjugating $$L$$ to $$H$$. Let $$Q$$ be the conjugate of $$P$$ by $$g$$. Then, $$N_K(Q) = H$$: This follows from the fact that conjugation, being an automorphism, preserves normalizers.
 * 7) $$Q$$ is a $$H$$-invariant $$p$$-Sylow subgroup of $$G$$: This is an immediate corollary of the preceding step.