Frattini subgroup of finite group is quotient-coprime automorphism-faithful

Verbal statement
The fact about::Frattini subgroup of a finite group is a fact about::quotient-coprime automorphism-faithful subgroup.

Statement with symbols
Let $$G$$ be a finite group, and let $$\Phi(G)$$ denote the Frattini subgroup of $$G$$. If $$\varphi$$ is a non-identity automorphism of $$G$$ such that the order of $$\varphi$$ is relatively prime to the order of $$G$$, then the induced map by $$\varphi$$ on $$G/\Phi(G)$$ is also a non-identity automorphism. In other words, the kernel of the natural map:

$$\operatorname{Aut}(G) \to \operatorname{Aut}(G/\Phi(G))$$

has no prime factors of its order other than the prime factors of the order of $$G$$.

Related facts

 * Burnside's theorem on coprime automorphisms and Frattini subgroup: The special case of this where $$G$$ is of prime power order.
 * Stability group of subnormal series of finite group has no other prime factors
 * Centralizer of coprime automorphism in homomorphic image equals image of centralizer

Facts used

 * 1) uses::Frattini subgroup is characteristic
 * 2) uses::Characteristic implies normal
 * 3) uses::Centralizer of coprime automorphism in homomorphic image equals image of centralizer: If $$\varphi$$ is an automorphism of a finite group $$G$$ of order coprime to $$G$$, and if $$N$$ is a normal $$\varphi$$-invariant subgroup of $$G$$, with $$\pi:G \to G/N$$ being the quotient map, then $$\pi(C_G(\varphi)) = C_{G/N}(\varphi)$$. In other words, every coset of $$N$$ that is centralized by $$\varphi$$ contains an element of $$G$$ that is centralized by $$\varphi$$.
 * 4) uses::Equivalence of definitions of Frattini subgroup: This states that the Frattini subgroup equals the set of nongenerators: any element of the Frattini subgroup can be discarded from a generating set while still keeping it a generating set.

Proof
Given: A finite group $$G$$, an automorphism $$\varphi$$ of $$G$$ whose order is relatively prime to the order of $$G$$. Further $$\varphi$$ induces the identity automorphism on $$G/\Phi(G)$$.

To prove: $$\varphi$$ is the identity automorphism.

Proof:


 * 1) $$\Phi(G)$$ is a normal $$\varphi$$-invariant subgroup of $$G$$: By fact (1), $$\Phi(G)$$ is characteristic, hence it is both normal (fact (2)) and invariant under $$\varphi$$.
 * 2) Let $$\pi:G \to G/\Phi(G)$$ be the quotient map. Then, $$\pi(C_G(\varphi)) = G/\Phi(G)$$: All the conditions holds for fact (3), so $$\pi(C_G(\varphi)) = C_{G/\Phi(G)}(\varphi)$$. By the given data, every element of $$G/\Phi(G)$$ is fixed under $$\varphi$$, so the right side is $$G/\Phi(G)$$.
 * 3) $$\Phi(G)C_G(\varphi)= G$$: Since $$\pi(C_G(\varphi)) = G/\Phi(G)$$, $$C_G(\varphi)$$ intersects every coset of $$\Phi(G)$$ in $$G$$, hence its product with $$\Phi(G)$$ equals $$G$$.
 * 4) $$C_G(\varphi) = G$$: The elements of $$\Phi(G)$$, along with the elements of $$C_G(\varphi)$$, together generate $$G$$. By fact (4), the elements of $$\Phi(G)$$ can be thrown out one by one, preserving at every stage the property of generating $$G$$. Since $$G$$ is finite, $$\Phi(G)$$ is finite, so all its elements can be removed, yielding that $$\langle C_G(\varphi) \rangle = G$$. Since $$C_G(\varphi)$$ is a subgroup, we get $$C_G(\varphi) = G$$.