# Equivalence of definitions of subgroup-conjugating automorphism

This article gives a proof/explanation of the equivalence of multiple definitions for the term subgroup-conjugating automorphism
View a complete list of pages giving proofs of equivalence of definitions

## The definitions that we have to prove as equivalent

### Subgroup-conjugating automorphism

An automorphism $\sigma$ of a group $G$ is termed a subgroup-conjugating automorphism if, for any subgroup $H$ of $G$, $H$ and $\sigma(H)$ are conjugate subgroups.

### Permutation-extensible automorphism

An automorphism $\sigma$ of a group $G$ is termed a permutation-extensible automorphism if, for any injective homomorphism $i: H \to \operatorname{Sym}(S)$, where $\operatorname{Sym}(S)$ is the symmetric group on a set, there is a permutation $\alpha$ of $S$ such that $i \circ \sigma = c_\alpha \circ i$, where $c_\alpha$ denotes conjugation by $\alpha$.

### Permutation-pushforwardable automorphism

An automorphism $\sigma$ of a group $G$ is termed a permutation-extensible automorphism if, for any homomorphism (not necessarily injective) $\rho: H \to \operatorname{Sym}(S)$, where $\operatorname{Sym}(S)$ is the symmetric group on a set, there is a permutation $\alpha$ of $S$ such that $\rho \circ \sigma = c_\alpha \circ \rho$, where $c_\alpha$ denotes conjugation by $\alpha$.

## Proof

### Permutation-extensible implies subgroup-conjugating

In this proof, we use the notation $c_g$ for conjugation by $g$, which is the map $x \mapsto gxg^{-1}$.

Given: A group $G$, a subgroup $H$, a permutation-extensible automorphism $\sigma$ of $G$.

To prove: $\sigma(H)$ is a conjugate subgroup to $H$.

Proof: The case that $H$ is the trivial subgroup is obvious, so we give the proof for $H$ nontrivial.

Let $S = G/H$ and $T = G$, with $G$ acting on $S$ and $T$ both by left multiplication. Let $U$ be the disjoint union of $S$ and $T$. $G$ acts faithfully on $U$, so we have an embedding:

$G \to \operatorname{Sym}(U)$.

By the condition, there exists $\alpha \in \operatorname{Sym}(U)$ such that $\sigma$ extends to conjugation by $\alpha$ in $\operatorname{Sym}(U)$. Consider the element $H \in U$. Clearly, the isotropy subgroup of the element $\alpha H \in U$ is the subgroup $c_\alpha(H)$ in $G$, which equals $\sigma(H)$.

Now, observe that $\alpha H$ cannot be in $T$, because then its isotropy group would be trivial, and $\sigma(H)$ cannot be trivial if $H$ is nontrivial. Thus, $\alpha H \in S$, so there exists $g \in G$ such that $\alpha H = gH$. Thus, the isotropy subgroup of $\alpha(H)$ is the conjugate subgroup $c_g(H)$, and thus $\sigma(H) = c_g(H)$.

### Subgroup-conjugating implies permutation-pushforwardable

Given: A group $G$, an automorphism $\sigma$ of $G$ such that $\sigma$ sends every subgroup to a conjugate subgroup. A homomorphism $\rho:G \to \operatorname{Sym}(S)$ for some set $S$.

To prove: There exists a permutation $\alpha$ of $S$ such that $\rho \circ \sigma = c_\alpha \circ \rho$.

Proof: Let $\mathcal{O}$ be the orbit of some point in $S$ under the induced action of $G$. Let $x \in \mathcal{O}$, and let $H$ be the isotropy subgroup of $x$. Let $x'$ be any point in $\mathcal{O}$ whose isotropy subgroup is $\sigma(H)$. Such a point exists because $H$ and $\sigma(H)$ are conjugate subgroups. Now define:

$\alpha(g \cdot x) = \sigma(g) \cdot x'$.

This is well-defined and gives a permutation of the orbit $\mathcal{O}$. If we define $\alpha$ in this way for each orbit, we get a permutation of $S$ and it satisfies the condition $\rho \circ \sigma = c_\alpha \circ \rho$.

### Puermutation-pushforwardable implies permutation-extensible

This implication is obvious.