Normalizer of subnormal subgroup. Problem 2.A.9 from FGT by Isaacs I try to solve problem 2.A.9 from M. Isaacs Finite Group Theory (FGT). I give it here.

Let $G$ be a finite group and $\pi$ be a set of primes. Let $H$ be a subnormal subgroup of $G$ and assume that
  $H=O^{\pi}(H)$, where $\pi$ is the set of primes. Show that
  $O_{\pi}(G)$ normalizes $H$.

I explain the notations.
$O^{\pi}(H)$ is the smallest normal subgroup of $H$ such that $H/O^{\pi}(H)$ is a $\pi$-group (that is every prime, which divide the order of the group $H/O^{\pi}(H)$) is contained in $\pi$).
$O_{\pi}(G)$ is the largest normal  $\pi$-subgroup of $G$.
I try to use the standard approach using induction on $|G|$. But this does not lead to success. Maybe there is another approach in this problem?
 A: Work by induction on $|G|$. As in the given hint assume that $G=HO_{\pi}(G)$. Indeed if this isn't true then by the inductive hypothesis in the group $HO_{\pi}(G)$ we have that $O_{\pi}(G)$ normalizes $H$ .
Now we have that $H \lhd \lhd G$ and $[G:H]$ is a $\pi$ number. So by problem 2A.2 we have that $O^{\pi}(G) \le H$. But now $O^{\pi}(G) \lhd H$ and $H/O^{\pi}(G)$ is a $\pi$ group, which implies that $O^{\pi}(G) = O^{\pi}(H) = H$. Now obviously $H = O^{\pi}(G) \lhd G$ and so $O_{\pi}(G)$ normalizes $H$ .

[UPDATE]: Solution of the problem 2A.2 in the book.

2A.2 If $K \lhd \lhd G$ and $[G:K]$ is a $\pi$ number, show that $O^{\pi}(G) \le K$.

Work by induction on $|G|$. The base case is trivial, as well as the case when $K=G$. So assume that $K < G$. Then there exists $N \lhd G$, s.t. $K \lhd \lhd N$. Also $[N:K]$ is a $\pi$ number, so as $N$ is a proper subgroup of $G$, by the inductive hypothesis we have that $O^{\pi}(N) \le K$.
Now $O^{\pi}(N) \text{ char } N \lhd G \implies O^{\pi}(N) \lhd G$. We know that $[G:N]$ is a $\pi$ number so $|G/O^{\pi}(N)| = |G/N||N/O^{\pi}(N)|$, which is a $\pi$ number, so $G/O^{\pi}(N)$ is a $\pi$-group. But we know that $O^{\pi}(G)$ is contained in all normal subgroups of $G$, whose factor groups are $\pi$-groups, so $O^{\pi}(G) \le O^{\pi}(N) \le K$ and hence the proof. 
