Frattini's argument

For automorph-conjugate subgroups
Let $$H$$ be a normal subgroup of $$G$$ and $$P$$ an automorph-conjugate subgroup of $$H$$. Then:

$$HN_G(P) = G$$

where $$N_G(P)$$ denotes the normalizer of $$P$$ in $$G$$.

For Sylow subgroups
Let $$H$$ be a normal subgroup of $$G$$ and $$P$$ a fact about::Sylow subgroup of $$H$$. Then:

$$HN_G(P) = G$$

where $$N_G(P)$$ denotes the normalizer of $$P$$ in $$G$$.

For characteristic subgroups of Sylow subgroups
Let $$H$$ be a normal subgroup of $$G$$, $$P$$ be a Sylow subgroup of $$H$$, and $$K$$ be a characteristic subgroup of $$P$$. In other words, $$K$$ is a fact about::characteristic subgroup of Sylow subgroup of $$H$$. Then:

$$HN_G(K) = G$$.

Facts used

 * 1) uses::Sylow implies automorph-conjugate
 * 2) uses::Characteristic implies automorph-conjugate
 * 3) uses::Automorph-conjugacy is transitive

Proof for automorph-conjugate subgroups
(This proof uses the left action convention)

Given: $$H$$ a normal subgroup of $$G$$. $$P$$ an automorph-conjugate subgroup of $$H$$.

To prove: $$HN_G(P) = G$$.

Proof: Let $$g \in G$$. Consider $$gPg^{-1}$$. Since $$H$$ is normal, the map $$x \mapsto gxg^{-1}$$ is an automorphism restricted to $$H$$. Since $$P$$ is automorph-conjugate in $$H$$, there exists $$h \in H$$ such that $$hPh^{-1} = gPg^{-1}$$.

Then, $$g^{-1}h = x\in N_G(P)$$, and hence $$g = hx^{-1}$$, with $$h \in H, x^{-1} \in N_G(P)$$. thus, every element of $$G$$ can be expressed as the product of an element of $$H$$ and an element of $$N_G(P)$$, and we are done.

Proof for Sylow subgroups
This follows from the statement for automorph-conjugate subgroups and fact (1).

Proof for characteristic subgroups
By facts (1), (2) and (4), every characteristic subgroup of a Sylow subgroup is automorph-conjugate, so this statement again follows from the statement for automorph-conjugate subgroups.