The commutator-intersection conjecture for RCC loop folders

Let $G$ be a finite group and let $H\leq G$ be a subgroup. A subset $T\subseteq G$ is a transversal for $H\backslash G$ if it contains exactly one representative of each right coset, and it is $G$-invariant when it is invariant under the action of $G$; assume that $1\in T$. Write $G'$ for the commutator subgroup of $G$.

Commutator-intersection conjecture. If $H$ is abelian and $(G,H,T)$ is an RCC loop folder, then

$$G'\cap H=\{1\}.$$

This is proposed as the converse to the sufficient condition previously established in the paper: $G'\cap H=\{1\}$ guarantees a $G$-invariant transversal containing $1$. The dihedral example shows that the existence of such a transversal alone does not force $H$ to be abelian; the conjecture asserts that, once $H$ is abelian, the commutator intersection must be trivial.