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 $\mathrm{I}\mathrm{n}\mathrm{n}(G)$, and automorphism group $\mathrm{A}\mathrm{u}\mathrm{t}(G)$. Suppose the following two statements are true:

• Every automorphism of $G$ is a center-fixing automorphism: it fixes every element of the center $Z(G)$.
• $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ is a maximal subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(G)$.

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 ${\sigma }^{\prime }$ 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. 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. Equivalence of definitions of central factor

Proof

Given: A group $G$ such that every automorphism of $G$ fixes every element of $Z(G)$, and $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ is a maximal subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(G)$.

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 $\alpha :K\to \mathrm{A}\mathrm{u}\mathrm{t}(G)$ 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 $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ or $\mathrm{A}\mathrm{u}\mathrm{t}(G)$: The image of $\alpha$ is a subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(G)$. It contains $\mathrm{I}\mathrm{n}\mathrm{n}(G)$, because $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ equals $\alpha (G)$. Since $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ is a maximal subgroup of $\mathrm{A}\mathrm{u}\mathrm{t}(G)$, the image is either $\mathrm{I}\mathrm{n}\mathrm{n}(G)$ or $\mathrm{A}\mathrm{u}\mathrm{t}(G)$.
2. If the image is $\mathrm{A}\mathrm{u}\mathrm{t}(G)$, 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 $\mathrm{I}\mathrm{n}\mathrm{n}(G)$, 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 $K$.