Let $H\subseteq G$ with $M\lhd H$ and $H\cap H^x\subseteq M$ when $x\notin H$. Show there exists $N\lhd G$ with $NH=G$ & $N\cap H=M$

Question

I'm struggling to prove Question 7.2. from Isaacs' Character Theory of Finite groups.

The question is:

Let $H\subseteq G$ with $M\triangleleft H$ and suppose that $H\cap H^{x}\subseteq M$ whenever $x\notin H$. Show that there exists $N\triangleleft G$ with $NH=G$ and $N\cap H=M$.

I know that I need to mimic the proof of Frobenius' theorem. But I don't know exactly at what point I need to make changes and which changes.

I was wondering if to define $N=G\setminus (\bigcup_{x\in G\setminus H}xHx^{-1})\cup M$ could help to solve this question.

I had the idea to prove it considering the class functions on $H$ that vanish on $M$, but I'm not sure at which point should make the distinction on the proof of Frobenius' theorem.

Answer 1

In addition to what Derek Holt hinted: you want to show that $N=G - \bigcup _{x \in G}(H - M)^x$ is in fact a normal subgroup. Mimicking the proof of Theorem 7.2 gets you $N=\bigcap_{\varphi \in Irr(H)}\{ker(\varphi^*): M \subseteq ker(\varphi)\}$. You also need to generalise the counting argument of Lemma (7.3).