Every automorphism is center-fixing and inner automorphism group is maximal in automorphism group implies every automorphism is normal-extensible

Statement
Suppose $$G$$ is a group with center $$Z(G)$$, inner automorphism group $$\operatorname{Inn}(G)$$, and automorphism group $$\operatorname{Aut}(G)$$. Suppose the following two statements are true:


 * Every automorphism of $$G$$ is a center-fixing automorphism: it fixes every element of the center $$Z(G)$$.
 * $$\operatorname{Inn}(G)$$ is a maximal subgroup of $$\operatorname{Aut}(G)$$.

Then, every automorphism of $$G$$ is a normal-extensible automorphism: whenever $$G$$ is a normal subgroup of some group $$K$$ and $$\sigma$$ is an automorphism of $$G$$, $$\sigma$$ extends to an automorphism $$\sigma'$$ of $$K$$.

Related facts

 * Centerless and maximal in automorphism group implies every automorphism is normal-extensible
 * Finite p-group with center of prime order and inner automorphism group maximal in p-Sylow-closure of automorphism group implies every p-automorphism is p-normal-extensible
 * Every automorphism is center-fixing and outer automorphism group is rank one p-group implies not every normal-extensible automorphism is inner

Facts used

 * 1) uses::Center-fixing implies central factor-extensible: If $$\sigma$$ is a center-fixing automorphism of a central factor $$G$$ of a group $$K$$, $$\sigma$$ extends to an automorphism of $$K$$.
 * 2) uses::Equivalence of definitions of central factor

Proof
Given: A group $$G$$ such that every automorphism of $$G$$ fixes every element of $$Z(G)$$, and $$\operatorname{Inn}(G)$$ is a maximal subgroup of $$\operatorname{Aut}(G)$$.

To prove: For any automorphism $$\sigma$$ of $$G$$, and any group $$K$$ containing $$G$$ as a normal subgroup, $$\sigma$$ extends to an automorphism of $$K$$.

Proof: Let $$\alpha:K \to \operatorname{Aut}(G)$$ be the homomorphism given by the action of $$K$$ on $$G$$ by conjugation. This map exists because $$G$$ is normal in $$K$$.


 * 1) The image of $$\alpha$$ is either $$\operatorname{Inn}(G)$$ or $$\operatorname{Aut}(G)$$: The image of $$\alpha$$ is a subgroup of $$\operatorname{Aut}(G)$$. It contains $$\operatorname{Inn}(G)$$, because $$\operatorname{Inn}(G)$$ equals $$\alpha(G)$$. Since $$\operatorname{Inn}(G)$$ is a maximal subgroup of $$\operatorname{Aut}(G)$$, the image is either $$\operatorname{Inn}(G)$$ or $$\operatorname{Aut}(G)$$.
 * 2) If the image is $$\operatorname{Aut}(G)$$, then every automorphism of $$G$$ extends to an inner automorphism of $$K$$, and we are done. (We say in this case that $$G$$ is a normal fully normalized subgroup of $$K$$).
 * 3) If the image is $$\operatorname{Inn}(G)$$, then $$G$$ is a central factor of $$K$$, i.e., $$K = G C_K(G)$$ (see fact (2)). Thus, by fact (1), any automorphism $$\sigma$$ that fixes the center of $$G$$ extends to an automorphism of $$K$$. Since we assumed that every automorphism fixes the center of $$G$$, we obtain that every automorphism of $$G$$ extends to an automorphism of $$K$$.