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Let $G$ be a finite group and $N$ be a normal subgroup of $G$ such that $G/N$ is nilpotent.

a) Prove that there exists a nilpotent subgroup $K$ of $G$ such that $G=NK.$

b) Suppose that $N$ is abelian and $Z\left(G\right) =\lbrace e \rbrace.$ Show that if $K$ is nilpotent and $G=NK$ then $K=N_G\left(K\right)$ and $K \cap N =\lbrace e \rbrace.$

Could anyone help me to give a solution for the problem! thank you!

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    $\begingroup$ For a), start with $K=G$. Then repeat following until $K$ is nilpotent. Choose $P \in {\rm Syl}_p(G)$ not normal in $K$. Use the Frattini Argument to show that $K = (N \cap K)N_K(P)$ (and hence $G=NK$) and replace $K$ by $N_K(P)$. $\endgroup$ – Derek Holt Nov 29 '16 at 8:29
  • $\begingroup$ The last equality in (b) cannot be what you wrote. Perhaps you meant $\;K\color{red}\cap N=\{e\}\;$ ? $\endgroup$ – DonAntonio Nov 29 '16 at 9:33
  • $\begingroup$ I'm sorry. I have just fixed it. thank you! $\endgroup$ – Hao Tran Nov 29 '16 at 12:25
  • $\begingroup$ could you propose a argument for (b)? tks $\endgroup$ – Hao Tran Nov 29 '16 at 12:28
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    $\begingroup$ If $K \cap N \ne \{ e \}$ then $K \cap N \cap Z(K)$ is nontrivial and lies in the centre of $G$. $\endgroup$ – Derek Holt Nov 29 '16 at 12:34
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Hao, here is the proof spelled out. Credits go to prof. Holt.

(a) We use induction on $|G|$. If $G$ is nilpotent there is nothing to prove (take $K=G$). If $G$ is not nilpotent take a Sylow $p$-subgroup $P$, that is not normal (remember that a finite group is nilpotent iff all its Sylow subgroups are normal). Hence, $N_G(P) \subsetneq G$. We are going to apply the induction step on $N_G(P)$: $PN/N \in Syl_p(G/N)$, whence $PN \unlhd G$, since $G/N$ is nilpotent. But $P \in Syl_p(G)$, so certainly $P \in Syl_p(PN)$. Apply the Frattini Argument: $G=N_G(P)PN=N_G(P)N$.
This implies that $G/N \cong N_G(P)N/N \cong N_G(P)/(N_G(P) \cap N)=N_G(P)/N_N(P)$, which is nilpotent. Now the induction step: it follows that we can find a nilpotent subgroup $K \leq N_G(P)$ with $N_G(P)=KN_N(P)$. We are going to apply Dedekind's Modular Law. $N_G(P)=KN_N(P)=K(N \cap N_G(P))=KN \cap N_G(P)$. This is equivalent to $N_G(P) \subseteq KN$. But we had $G=N_G(P)N$, hence certainly $G=KN$, and we are done.

There is another way to prove this, but I will not spell out all the details: if $\mathcal{K}=\{K \leq G: G=KN \}$, then this set is non-empty: $G \in \mathcal{K}$. Hence we can pick a minimal element from this set, say $K$. This $K$ turns out to be nilpotent. This is done by showing that $K \cap N \subseteq \Phi(K)$, the Frattini-subgroup of $K$. Since $K/(K \cap N) \cong G/N$ is nilpotent, you then can conclude that $K$ is in fact nilpotent.

(b) Here you need that a non-trivial normal subgroup of a nilpotent group has non-trivial intersection with the center of the whole group, see here. Assume that $K \cap N \neq 1$, that is, $K \cap N$ is a non-trivial normal subgroup of the nilpotent group $K$ and hence $K \cap N \cap Z(K) = N \cap Z(K) \neq 1$. But clearly, since $G=KN$, $N \cap Z(K) \subseteq Z(G)$, contradicting $Z(G)=1$. We conclude that $K \cap N=1$.

Now let’s study $N_G(K)$. Observe that $K \unlhd N_G(K)$ by definition and $N \cap N_G(K)=N_N(K) \unlhd N_G(K)$, since $N$ is normal. Also, again using Dedekind's Modular Law and that $G=KN$, we see that $N_G(K)=N_N(K)K$. Using $N_N(K) \cap K=1$ we arrive at $N_G(K)=N_N(K) \times K$. This implies that $N_N(K)$ centralizes $K$. But being a subgroup of the abelian group $N$, $N_N(K)$ also centralizes $N$. Hence, $N_N(K) \subseteq Z(G)=1$, and we are done.

Note The group $K$ here is an example of a so-called Carter subgroup of $G$: a self-normalizing nilpotent subgroup. Carter proved in 1961 that every solvable group has Carter-subgroups. In addition, the Carter subgroups are conjugate subgroups and thus isomorphic. Gaschütz (1963) viewed the Carter subgroups as analogues of Sylow subgroups and Hall subgroups, and unified their treatment with the theory of formations.

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    $\begingroup$ @Hoa Tran, you are welcome. If you deem my answer being the right one, please tick it as such (a green tick appears), as is the habit at this site. $\endgroup$ – Nicky Hekster Dec 9 '16 at 6:58
  • $\begingroup$ yeah! I did it! thank you! $\endgroup$ – Hao Tran Dec 29 '16 at 5:33

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