  Let $G$ be a finite group and $H$ a core-free subgroup of $G$. We will show
that if there exists a solvable, generating transversal of $H$ in $G$, then $G$
is a solvable group. Further, if $S$ is a generating transversal of $H$ in $G$
and $S$ has order 2 invariant sub right loop $T$ such that the quotient $S/T$
is a group. Then $H$ is an elementary abelian 2-group.
