When a simple group normalizes a subnormal subgroup of a finite group Out of curiosity I am working my way through Isaacs's Finite Group Theory and am stuck on problem 2A.4:

Let $(G,*)$ be a finite group with simple subgroup $N$ 
    such that $\forall H \lhd\lhd G: N H = H N$.
    Then, $\forall H \lhd\lhd G: N \subseteq N_G(H)$.

Notation $\lhd\lhd$ means being a subnormal subgroup; $N_G(H)$ denotes the normalizer of $H$ in $G$; $C_G(H)$ denotes the centralizer of $H$ in $G$.
My approach: strong induction on $|G|$.
Let $H \lhd\lhd G$.
Case 1 ($H$ is normal in $G$): Hence, $N_G(H) = G$ and the claim is evident via $N \subseteq G$.
Case 2 ($H$ is not normal in $G)$): Choose the penultimate term $U$ in a subnormal series, that is, $H \lhd\lhd U \lhd G$ where $H \subsetneq U \subsetneq G$.
Since $U \lhd G$, we know $U \cap N \lhd N$, that is, either $U \cap N = \{1\}$ or $U \cap N = N$ because $N$ is simple.
Case 2.1 ($U \cap N = N$): Then, $N \subseteq U$.
If $W \lhd\lhd U$, then $W \lhd\lhd U \lhd G$ and so $W \lhd\lhd G$, that is, $W N = N W$ by basis premiss.
Thus, $N \subseteq N_U(H) = U \cap N_G(H) \subseteq N_G(H)$ by induction hypothesis.
Case 2.2 ($U \cap N = \{1\}$): If $N \lhd G$, then $N \subseteq C_G(U) \subseteq C_G(H) \subseteq N_G(H)$.
If $N$ is not normal in $G$, then ???
Question: How can I remedy case 2.2? Or is my attempt simply inadequate? 
Thank you very much for your thoughts!
 A: An alternate answer (without using induction):
Let $H=H_0 \lhd H_1 … \lhd H_n=G$ be a subnormal series of $G$. If all $H_i$ contains N, there is nothing to prove.
Thus, we can assume there exists $H_i$ that does not contain $N$ such that $H_{i+1}$ does contain $N$. Then we get that $H_i \cap N$ is normal in $N$, and hence
$$H_i\cap N=1.$$
Now consider $H^N$ ($N$ closure of $H$). Since $HN$ is a group, we have $H^N \leq HN$. On the other hand, $H^N \leq H_i$ as $N$ normalizes $H_i$.
Thus, $H^N \leq  H_i \cap HN = H(H_i\cap N)=H$ by using Dedekind rule, which completes the proof.
A: First recall that $HN=NH$ is equivalent to say that $HN$ is a Group.
Claim: Let $H,N\leq G$, and suppose that $H$ is subnormal in $G$ and $N$ is simple.  If $HN$ is a group, then $N$ normalizes $H.$
Now let's proceed by induction on the order of $G$.
If $|HN|<|G|$, then the claim holds for the group $HN$ by induction. (Notice that $H\lhd \lhd HN$ also and $N$ is still simple!)
Thus, $G=HN$. On the other hand, $H\cap N \lhd \lhd N$ (why?). Due the simplicity of $N$, we get $H\cap N=N$ or $H\cap N=1$.  If the former case holds, the claim is naturally true.
Now suppose that $H\cap N=1$. Let $H\lhd U\lhd \lhd G$. (That is $U\neq H$). In this case, notice that $U\cap N\neq 1$. This forces that $N\subseteq U$, and so $U=G$. As a consequence, $H\lhd G$ which completes the proof.
I hope this helps.
