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!

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

I hope this helps.

  • $\begingroup$ Thank you very much your time! I could solve my problem with your ideas! $\endgroup$
    – Moritz
    Oct 7 '18 at 14:14
  • $\begingroup$ You are welcome :) $\endgroup$
    – mesel
    Oct 7 '18 at 14:15

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