Normal and centralizer-free implies automorphism-faithful

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., centralizer-free normal subgroup) must also satisfy the second subgroup property (i.e., automorphism-faithful normal subgroup)
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Statement

Suppose H is a subgroup of a group G such that:

  • H is centralizer-free in G: The centralizer C_G(H) is trivial.
  • H is normal in G.

Then, H is an Automorphism-faithful subgroup (?) of G: every nontrivial automorphism of G that restricts to H must restrict to a nontrivial automorphism of H.

Related facts

Analogues in other algebraic structures

Facts used

  1. Three subgroup lemma

Proof

Hands-on proof

Given: A group G, a normal subgroup H such that C_G(H) is trivial. An automorphism \sigma of G that acts as the identity on H.

To prove: \sigma is the identity automorphism.

Proof: Note that for g \in G, h \in H, we have, by normality, that ghg^{-1} \in H. Thus, since \sigma acts as the identity on H, we get:

\sigma(ghg^{-1}) = ghg^{-1}.

Expand the left side and use that \sigma(h) = h to obtain:

\sigma(g)h\sigma(g)^{-1} = ghg^{-1}.

Rearranging this yields that g^{-1}\sigma(g) commutes with h. Since this holds true for every h \in H, we obtain that g^{-1}\sigma(g) \in C_G(H). Since C_G(H) is trivial, we obtain that for every g \in G, g^{-1}\sigma(g) is the identity element, yielding g = \sigma(g) for all g \in G. This yields that \sigma is the identity automorphism.

Proof using three subgroup lemma

Note that although this proof may superficially appear different from the preceding one, it is a less hands-on formulation of the same proof. This style of proof may be more useful when proving similar statements that are more complicated; for instance: normal and self-centralizing implies coprime automorphism-faithful.

Given: A group G, a normal subgroup H such that C_G(H) is trivial. An automorphism \sigma of G that acts as identity on H.

To prove: \sigma is the identity automorphism.

Proof: Suppose A is the subgroup generated by \sigma in \operatorname{Aut}(G), and let's assume we're working in G \rtimes A. By assumption, [H,A] is trivial. Since H is normal in G, we have [G,H] \le H. Thus, we have:

  • [[G,H],A] \le [H.A], which is trivial, so [[G,H],A] is trivial
  • [[H,A],G] is trivial, because [H,A] is trivial

So, by fact (1) (the three subgroup lemma), [[G,A],H] is trivial. Since [G,A] \le G, we see that [G,A] \le C_G(H), so by the assumption of C_G(H) being trivial, we obtain that [G,A] is trivial. This translates to saying that every element of A fixes every element of G, so \sigma must be the identity automorphism.