Base of a wreath product implies subset-conjugacy-closed

Statement with symbols
Suppose $$G$$ is the internal wreath product of subgroups $$H$$ and $$K$$. In other words, $$H$$ is the base of a wreath product in $$G$$. Then, $$H$$ is a subset-conjugacy-closed subgroup of $$G$$: if $$A, B$$ are subsets of $$H$$ and there is a $$g \in G$$ such that $$gAg^{-1} = B$$, then there exists a $$h \in H$$ such that $$hah^{-1} = gag^{-1}$$ for all $$a \in A$$.

Other facts about subset-conjugacy-closed

 * Direct factor implies subset-conjugacy-closed
 * Retract implies subset-conjugacy-closed
 * Central factor implies subset-conjugacy-closed

Closely related facts about base of a wreath product

 * Base of a wreath product satisfies intermediate subgroup condition
 * Base of a wreath product implies conjugacy-closed
 * Base of a wreath product implies 2-subnormal
 * Base of a wreath product implies right-transitively 2-subnormal

Proof
Given: A group $$G$$ that is the internal wreath product of subgroups $$H$$ and $$G$$.

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

Proof: The statement is trivial if $$A, B$$ contain no non-identity element, so we assume without loss of generality that $$A$$ contains a non-identity element.

We have $$G = (H \times H \times \dots \times H) \rtimes K$$. Let $$L$$ be the subgroup $$H \times H \times \dots H$$. Then, $$G = KL$$, and the subgroup $$H$$ we are studying is the first direct factor of $$L$$.


 * 1) We can write $$g = kl$$, where $$k \in K$$ and $$l \in L$$, with $$l = (h_1, h_2, \dots, h_n)$$.
 * 2) If $$c_g$$, $$c_k$$, and $$c_l$$ denote conjugation by $$g,k,l$$ respectively, then math>c_g = c_k \circ c_l.
 * 3) The restriction of $$c_l$$ to $$H$$ is equal to conjugation by $$h_1$$. Thus, restricted to $$H$$, we have $$c_g = c_k \circ c_{h_1}$$.
 * 4) $$c_k$$ permutes the direct factors of $$L$$. But we know that $$c_k$$ sends at least one non-identity element of $$H$$ to within $$H$$. This forces the action of $$c_k$$ on the direct factors of $$L$$ to preserve the first direct factor; equivalently, the permutation induced by $$k$$ fixes the first element.
 * 5) Thus, restricted to $$H$$, $$c_k$$ is trivial and we get $$c_g = c_{h_1}$$ for the whole of $$H$$. In particular, $$c_g = c_{h_1}$$ when restricted to the subset $$A$$.