Normal and self-centralizing implies coprime 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., self-centralizing normal subgroup) must also satisfy the second subgroup property (i.e., coprime automorphism-faithful normal subgroup)
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This article states and (possibly) proves a fact about a finite group and a Coprime automorphism group (?): a subgroup of the automorphism group whose order is relatively prime to the order of the group itself.
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Statement

Verbal statement

Any normal Self-centralizing subgroup (?) of a finite group is coprime automorphism-faithful.

Statement with symbols

Suppose G is a finite group and H is a normal subgroup of G such that C_G(H) \le H (or equivalently, C_G(H) = Z(H)). Then, if \sigma is a non-identity automorphism of G whose order is relatively prime to the order of G, and if \sigma(H) = H, then the restriction of \sigma to H is a non-identity automorphism of H.

Related facts

Similar facts

Opposite facts

Facts used

  1. Three subgroup lemma
  2. Stability group of subnormal series of finite group has no other prime factors

Proof

Hands-on proof

Given: A finite group G, a normal subgroup H such that C_G(H) \le H. An automorphism \sigma of G that acts as identity on H, and such that the order of \sigma is relatively prime to the order of G

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) \le H by assumption, we obtain that g^{-1}\sigma(g) \in H, and thus, g and \sigma(g) are in the same coset of H in G. Thus, \sigma stabilizes the subnormal series:

\{ e \} \triangleleft H \triangleleft G.

Thus, by fact (2), \sigma must be the identity automorphism.

Proof using three subgroup lemma

Note that this proof superficially appears different from the preceding proof, but the part of the proof using the three subgroup lemma is precisely equivalent to the initial part of the proof above upto the point where we deduce that g^{-1}\sigma(g) \in H. Thinking of the proof in terms of the three subgroup lemma, rather than in terms of step-by-step manipulations as outlined above, is more useful in some situations.

Given: A finite group G, a normal subgroup H such that C_G(H) \le H. An automorphism \sigma of G that acts as identity on H, and such that the order of \sigma is relatively prime to the order of G

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 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 self-centralizing, we obtain that [G,A] \le H. Thus, A stabilizes the subnormal series:

\{ e \} \triangleleft H \triangleleft G

Now, using the fact that the stability group of the subnormal series of a finite group has no other prime factors, we conclude that A must be the trivial group, so \sigma is the identity automorphism.