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 $\alpha$.

Related facts

Applications

• Extensible implies subgroup-conjugating

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.