Normal and centralizer-free implies automorphism-faithful

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 fact about::automorphism-faithful subgroup of $$G$$: every nontrivial automorphism of $$G$$ that restricts to $$H$$ must restrict to a nontrivial automorphism of $$H$$.

Related facts

 * Normal and self-centralizing implies coprime automorphism-faithful

Analogues in other algebraic structures

 * Centralizer-free ideal implies automorphism-faithful
 * Centralizer-free ideal implies derivation-faithful

Facts used

 * 1) uses::Three subgroup lemma

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.