# Finite-quotient-pullbackable implies inner

This article gives the statement and possibly, proof, of an implication relation between two automorphism properties. That is, it states that every automorphism satisfying the first automorphism property (i.e., finite-quotient-pullbackable automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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## Statement

Suppose $H$ is a finite group and $\sigma$ is a finite-quotient-pullbackable automorphism of $G$: in other words, for any surjective homomorphism $\rho:G \to H$, there is an automorphism $\sigma'$ of $G$ such that $\rho \circ \sigma' = \sigma \circ \rho$.

Then, $\sigma$ is an inner automorphism of $G$.

## Facts used

1. Every finite group is the Fitting quotient of a p-dominated group for any prime p not dividing its order: Suppose $H$ is a finite group and $p$ is a prime not dividing the order of $H$. Then, there exists a finite complete group $G$ such that the Fitting subgroup $F(G)$ is a $p$-group, and $G$ is a semidirect product of $F(G)$ and $H$. In particular, $G/F(G) \cong H$.

## Proof

Given: A finite group $H$, an automorphism $\sigma$ of $H$ such that for any surjective homomorphism $\rho:G \to H$ there is an automorphism $\sigma'$ of $G$ such that $\rho \circ \sigma' = \sigma \circ \rho$.

To prove: $\sigma$ is inner.

Proof: Let $p$ be a prime not dividing the order of $H$. Using fact (1), there is a finite group $G$ such that the Fitting subgroup $F(G)$ is a $p$-group, and $G/F(G)$ is isomorphic to $H$. Let $\rho:G \to H$ be the quotient map.

By assumption, there exists an automorphism $\sigma'$ of $G$ such that $\rho \circ \sigma' = \sigma \circ \rho$. Since $G$ is complete, $\sigma'$ is inner, so there exists $g \in G$ such that $\sigma'$ is conjugation by $g$. It is easy to see then that $\sigma$ equals conjugation by the element $\rho(g)$, and thus, $\sigma$ is inner.