# Finite solvable-extensible 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 solvable-extensible automorphism) must also satisfy the second automorphism property (i.e., inner automorphism)
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## Statement

Suppose $H$ is a finite solvable group and $\sigma$ is a finite solvable-extensible automorphism: in other words, $\sigma$ extends to an automorphism of $G$ for any finite solvable group $G$ containing $H$. Then, $\sigma$ is an inner automorphism of $H$.

## 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 p-dominated group $G$ with $H$ as Fitting quotient: in other words, there exists a finite complete group $G$ such that the Fitting subgroup $F(G)$ is a $p$-group, and $H$ is a subgroup of $G$ such that $G = F(G) \rtimes H$.
2. Prime power order implies nilpotent, Nilpotent implies solvable
3. Solvability is extension-closed

## Proof

Given: A finite group $H$, a finite solvable-extensible automorphism $\sigma$ of $H$.

To prove: $\sigma$ is inner.

Proof: Let $p$ be a prime not dividing the order of $H$. Consider the group $G$ constructed by fact (1). Note first that $F(G)$ is a $p$-group, hence by fact (2), is solvable. Thus, both $F(G)$ and $G/F(G) \cong H$ are solvable, and hence, by fact (3), $G$ itself is a finite solvable group.

Since $\sigma$ is finite solvable-extensible, $\sigma$ extends to an automorphism $\sigma'$ of $G$. Further, since $G$ is complete, there exists $g \in G$ such that $\sigma'$ is conjugation by $g$.

Let $\rho:G \to H$ be the retraction with kernel $F(G)$. Note that conjugation by $g$ preserves $F(G)$, hence it induces a conjugation map on $H$ as a quotient, namely, conjugation by the element $\rho(g) \in H$. However, since the restriction of $\rho$ to the subgroup $H$ is the identity map, we conclude that conjugation by $g$ has the same effect on $H$ as conjugation by $\rho(g)$. In particular, $\sigma$ equals conjugation by $\rho(g)$, and hence is inner.