# 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!

• I could suggest you to follow "1-->3-->4-->5-->7-->6".Then you can proceed to any chapter you like. – mesel Oct 26 '18 at 17:09
• Thank you! I would not have expected this sequence of chapters. Any insight by an expert is greatly appreciated. – Moritz Oct 27 '18 at 19:56

First recall that $$HN=NH$$ is equivalent to say that $$HN$$ is a Group. 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.)
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.