Let $G$ be a finite group and $H$ be a subgroup of $G$. For a set of primes $\pi$ we define the $\pi-$closure of $H$ as $H^{\pi}=\langle H^g| g \:\text{is a}\: \pi-\text{element of } G\rangle$. When we can conclude that $H^{\pi}$ equals the normal closure of $H$ in $G$?
For example, this happens when all $\pi$-elements generates $G$. This condition is sufficient but not necessary.
Example: Set $G=A_4$ and suppose that $H=\langle (1 \:2)\rangle$. Then $|H|=2$ and $3-$closure of $H$ equals $H^G$.