So, I have been trying to understand if this is true for quite sometime.
Suppose you have a group $G$ and a subgroup $H < G$. Suppose also that $H$ is normal on its normal closure, i.e. $$H\vartriangleleft H^G \vartriangleleft G$$
$H$ is then a 2-subnormal subgroup of $G$ and I saw that all 2-subnormal subgroups are conjugate-permutable, i.e. $$ \forall g\in G\ ,\ H^g H = H H^g$$
My question: Does this mean that $$H^G = \prod_{g\in G} H^g\ \ ?$$
Thank you!