Does the Product of conjugates of some subgroups commutes with all elements of another subgroup

Question

Let $U,V \le G$ abelian subgroups such that $V$ is normal and $G$ be finite, does it hold that for $v \in V$ we have $$ u \cdot \left( \prod_{u' \in U} v^{u'} \right) = \left( \prod_{u' \in U} v^{u'} \right) \cdot u $$ for each $u \in U$?

Answer 1

This does hold. Recall that $zx^yz^{-1}=zyxy^{-1}z^{-1}=(zy)x(zy)^{-1}=x^{zy}$ and $x^yz^y=(xz)^y$. Therefore... $$u \cdot \left( \prod_{u' \in U} v^{u'} \right) \cdot u^{-1} = \prod_{u' \in U} v^{uu'} = \prod_{u' \in U} v^{u'}$$ The reason this last equality holds is that as $u'$ runs through $U$, the product $uu'$ runs through $U$ (left multiplication is 1-1 and onto). Thus we can re-index and get the desired result.

Edit: I originally said, you don't need most of the assumptions. It is true that $U$ doesn't need to be abelian (just a finite subgroup will do). However, $v$'s conjugates needs to commute each other (well, at least the conjugates you get when conjugating by $U$). But your assumption that $V$ is normal guarantees that $v$'s entire conjugacy class is contained in $V$ and $V$ is assumed to be abelian so $v$'s conjugates do commute with each other.

Actually, without the assumption $v$'s conjugates commute with each other, the product wouldn't even be well-defined!