Base of a wreath product implies subset-conjugacy-closed

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., base of a wreath product) must also satisfy the second subgroup property (i.e., subset-conjugacy-closed subgroup)
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Statement

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 $gA{g}^{-1}=B$, then there exists a $h\in H$ such that $ha{h}^{-1}=ga{g}^{-1}$ for all $a\in A$.

Definitions used

Base of a wreath product

Further information: Wreath product, Internal wreath product, base of a wreath product

Subset-conjugacy-closed subgroup

Further information: Subset-conjugacy-closed subgroup

Related facts

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 $gA{g}^{-1}=B$ for subsets $A,B\le H$ and $g\in G$, then there exists $h\in H$ such that $ha{h}^{-1}=ga{g}^{-1}$ 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 $G=(H\times H\times \dots \times H)⋊K$. 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</math>.
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 $A$.