Frattini's argument
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This article gives a proof/explanation of the equivalence of multiple definitions for the term automorph-conjugate subgroup
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Contents |
Statement
For automorph-conjugate subgroups
Let H be a normal subgroup of G and P an automorph-conjugate subgroup of H. Then:
HNG(P) = G
where NG(P) denotes the normalizer of P in G.
For Sylow subgroups
Let H be a normal subgroup of G and P a Sylow subgroup of H. Then:
HNG(P) = G
where NG(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 characteristic subgroup of Sylow subgroup of H. Then:
HNG(K) = G.
Facts used
- Sylow implies automorph-conjugate
- Characteristic implies automorph-conjugate
- Automorph-conjugacy is transitive
Proof
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: HNG(P) = G.
Proof: Let
. Consider gPg − 1. Since H is normal, the map
is an automorphism restricted to H. Since P is automorph-conjugate in H, there exists
such that hPh − 1 = gPg − 1.
Then,
, and hence g = hx − 1, with
. thus, every element of G can be expressed as the product of an element of H and an element of NG(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.