Is the normalizer of a subgroup of a maximal subgroup contained in that maximal subgroup?

Question

Let G be a finite group, H a maximal proper subgroup of G and K a subgroup of H. Is the normalizer of K in G, $N_GK$, a subgroup of H. Now $N_GK$ is certainly contained in some maximal subgroup, maybe more than one, but why is it contained in H. This should be really elementary.

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As is pointed out below, there are trivial counterexamples. K = {e} for example, or normal in G. So it es elementarily false. I will mark this answered.

Answer 1

There are several problems with what you write.

First: it is not true that $N_G(K)$ "is certainly contained in a maximal subgroup"; that's only true if $N_G(K)\neq G$, that is, if $K$ is not normal in $G$... a possibility you never excluded.

Second: It is not necessarily contained in $H$. Here is an example: take $G=S_3\times C_3$, let $H=S_3\times\{e\}$, and take $A$ to be a subgroup of $S_3$ of order $2$, and let $K$ be the subgroup given by $A\times\{e\}$. This is contained in $H$; but its normalizer is $A\times C_3$, which is not contained in $H$.

You have absolutely no reason to think that $N_G(K)$ will be contained in $H$ in general. If $K\triangleleft H$, then you know that $H\leq N_G(K)$ and hence that $N_G(K)=H$ or $G$, but I think that's about all you can say in this generality of setting.

Answer 2

There is one particular case where what you want is true and I think it is general enough that it is worth mentioning here. It is Theorem 9.2.13 in Derek Robinson's book "A Course in the Theory of Groups", p. 266.

Let $M$ be a nonnormal maximal subgroup of the finite soluble group $G$ and let $|G:M|=p^m$. If $Q$ is a Hall $p'$-subgroup of $M$, then $N_G(Q) \leq M$.