Equivalence of definitions of subset-conjugacy-closed subgroup

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

The definitions that we have to prove as equivalent

The following are equivalent for a subgroup $H$ of a group $G$:

1. For any subsets $A,B$ of $H$, such that there exists $g\in G$ with $gAg^{-1}=B$, there exists $h\in H$ such that $hah^{-1}=gag^{-1}$ for all $a\in A$.
2. $H$ is a subset-conjugacy-determined subgroup of itself with respect to $G$, i.e., if the fusion for subsets of $H$ in $G$, is contained in $H$.
3. $H$ possesses a Distinguished set of coset representatives (?) in $G$: In other words, there is a set $T$ of left coset representatives of $H$ in $G$ such that $hth^{-1}\in T$ for all $h\in H,t\in T$.

Proof

(1) and (2) are clearly the same statement, so we prove the equivalence between (1) and (3).

(3) implies (1)

Given: A subgroup $H$ of a group $G$ with a distinguished set $T$ of coset representatives.

To prove: If $A,B$ are subsets of $H$ and $g\in G$ is such that $gAg^{-1}=B$, then there exists $h\in H$ with $hah^{-1}=gag^{-1}$ for all $a\in A$.

Proof: Write $g=th$ for $t\in T,h\in H$. Note that this can be done because $T$ is a set of coset representatives for $H$ in $G$.

Then, suppose $a\in A$ and $gag^{-1}=b\in B$. We have:

$b=gag^{-1}=t(hah^{-1})t^{-1}$.

Let $a'=hah^{-1}$. Then, we get:

$b=ta't^{-1}$.

This can be rewritten as:

$a'=t^{-1}bt$.

Right-multiplying both sides by $b^{-1}$ yields:

$a'b^{-1}=t^{-1}btb^{-1}$.

The left side is in $H$. The right side is $t^{-1}(btb^{-1})$. Note that $btb^{-1}\in T$. For the ratio to be in $H$, we need that $t$ and $btb^{-1}$ are in the same coset of $H$, and since $T$ is a set ofcoset representatives, we get $btb^{-1}=t$. Thus, $t$ and $b$ commute. In particular, $a'=b$, so we get:

$b=hah^{-1}$.

Thus, for every $a\in A$, we get $gag^{-1}=hah^{-1}$, completing the proof.

(1) implies (3)

Given: A group $G$, a subset-conjugacy-closed subgroup $H$ of $G$.

To prove: $H$ has a distinguished set of coset representatives.

Proof: The proof involves the following two ideas:

• We can locally choose a representative for each coset, such that things do not mess up in that coset.
• If we pick a good coset representative in this sense, then all its conjugates are also good coset representatives.
• We can use the axiom of choice to pick out a distinguished set.

Here, $t$ is a good coset representative if whenever $a\in H$ is such that $tat^{-1}\in H$, then $tat^{-1}=a$.