Equivalence of definitions of subset-conjugacy-closed subgroup

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 fact about::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 $$tat^{-1} = a$$.