When $\pi$-closure equals the normal closure of a subgroup

Question

Let $G$ be a finite group and $\pi$ be a set of primes. For any subgroup $H$ of $G$ define the $\pi$-closure of $H$ as $H^{\pi}=\langle x^{-1}Hx| x\: \text{is a ${\pi}$-element of}\: G \rangle$. Can you help me to find a minimal $\pi$ such that the $\pi$-closure of $H$ equals the normal closure of $H$? Unfortunately, I have no idea to solve the problem.

Answer 1

Let $G=S_3$, $H=A_3$ and $\pi=\{3\}$. Then you can see that the $\pi$-closure of $H$ and the normal closure of $H$ are both equal to $H$.

If you take $K=\langle (1,2) \rangle$ and $\pi=\{2\}$, you also see that the $\pi$-closure of $K$ and the normal closure of $K$ are both equal to $G$.