and that $D_{s,\lambda}\cdot1_{G}=G$; hence $D_{s,\lambda}$ is a regular subgroup of ${\text{Perm}}(G)$. The verification that it is normalized by $ \lambda(G) $ is very similar to the verifications in and , using the fact that $\rho(G)$ and $\lambda(G)$ commute inside $ {\text{Perm}}(G) $. Similarly, $ D_{s,\rho} $ is a regular subgroup of $ {\text{Perm}}(G) $ that is isomorphic to $ D_{4} $ and normalized by $ \lambda(G) $. To show that the $ 3 $ choices of $ s $ yield $ 6 $ distinct groups, note that for each $ s $ the group $ D_{s,\lambda} $ is the only one that contains $ \lambda(s) $ and that $ D_{s,\rho} $ is the only one that contains $ \rho(s) $. Hence, by , the groups $D_{s,\lambda}$ and $D_{s,\rho}$ are the underlying groups of the 6 Hopf-Galois structures of type $D_4$.

For every regular subgroup $ N $ of $ {\text{Perm}}(G) $ corresponding to a Hopf-Galois structure on $ L/F $ we have $ \rho(\sigma^{2}) \in N $, and so $ Z(\rho(G)) \subseteq \rho(G) \cap N $. Clearly this is the case for $ N=\rho(G) $ and $ N=\lambda(G) $, and it is easy to verify that it holds for $ 