# Finite-quotient-pullbackable implies inner

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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.