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Having problems with the following statement:

If $N$ is a nontrivial normal subgroup of a nilpotent group $G$ then $N \cap Z(G) \neq \langle e \rangle$

Here $Z(G)$ denotes the center of $G$.

I've solved the finite case, where I reduced the problem to proving that a p-$group$ has this property. The infinite case resists my attempts though.

Since $G$ is nilpotent, the ascending central series ends with $G$ itself: $\langle e\rangle \subseteq Z_1(G) \subseteq \ldots \subseteq Z_n(G) = G$ where $Z_1(G) = Z(G)$. Since $N \subseteq G$, there must be a smallest $i$ such that $N \cap Z_i(G) \neq \langle e \rangle$. I would like to show that $i = 1$ in this case, or perhaps just show that $N \cap Z_i(G)$ commutes with elements of $G$. Could somebody give me a tip in the right direction?

Bartosz Malman
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    Consider $[N,G,G,\ldots]$. –  Apr 01 '12 at 19:47
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    This is called "Normality Large". There is a proof of this on Groupprops: http://groupprops.subwiki.org/wiki/Nilpotent_implies_center_is_normality-large – Alex Youcis Apr 01 '12 at 19:48
  • @alex, please excuse my ignorance. I'm reading the solution in the link. Why is $[G, Z_{i+1}(G)] \subset Z_i(G)$ ? I see that commutator of $Z_{i+1}(G)$ must be contained in $Z_{i}(G)$ as quotient $Z_{i+1}(G)/Z_{i}(G)$ is abelian, but why the entire $[G,Z_{i+1}(G)]$ ? – Bartosz Malman Apr 01 '12 at 20:25
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    @Barre: Because $Z_{i+1}$ is the inverse image in $G$ of the center of $G/Z_i$. In other words, every element of $Z_{i+1}$ commutes with every element of $G$, up to an element of $Z_i$, which is just another way of saying $[G, Z_{i+1}] \subset Z_i$. – Ted Apr 01 '12 at 20:42
  • @Ted oh, of course. In the quotient group $G/Z_i$ the subgroup $Z_{i+1}/Z_i$ is the center. But that implies for any $c \in Z_{i+1}$, $Z_icZ_ig = Z_icg = Z_igc$, and therefore $Z_i = Z_igcg^{-1}c^{-1}$ which means $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. So the all generators of $[G,Z_{i+1}]$ are contained in $Z_i$ and therefore the entire group is too. I think I should be able to get the rest of the proof from here, and will answer my own question later today or tomorrow, unless somebody beats me to it. Big thanks for everybody! – Bartosz Malman Apr 01 '12 at 20:56
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    Three comments: It is uncommon to use $C$ for the upper central series; usually one uses either $Z$ or $\zeta$. Second, in your argument, it is false that $[G,N\cap C_{i+1}]$ is empty; subgroups are never empty. Finally, you cannot conclude that $[G,N\cap C_{i+1}] = [G,N]\cap [G,C_{i+1}]$, though you can conclude $\subseteq$ instead of equality (which suffices). Otherwise, it looks fine. – Arturo Magidin Apr 01 '12 at 22:13
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    The "dual" of your argument (which I think is easier to understand): let $N_i=[N,G,G,\ldots,G]$, with $i$ $G$'s. Note $N_0=N$. Because $N$ is normal, $N_i\subset N$; because $G$ is nilpotent, for some $j$, $N_{j+1}=\lbrace1\rbrace$. Then $N_{j+1}=[N_j,G]=\lbrace1\rbrace$, so $N_j$ is central. –  Apr 01 '12 at 22:32
  • @Steve Indeed, this proof is much cleaner. I am not completely comfortable I understand it right, but actually Hungerford characterises nilpotent groups in a similar way in one of the exercises (nilpotent iff $\gamma_n(G) = \langle e \rangle$ for some $n$, where $\gamma_1(G) = G, \gamma_2(G) = (G,G), \gamma_i(G) = (\gamma_{i-1}(G), G)$. I will have to look into this, but is it correct that $N_i \subset \gamma_i(G)$ ? – Bartosz Malman Apr 02 '12 at 12:32
  • @BArre: Yes, since $N\le G$, we have $[N,G]\le [G,G]$, etc. –  Apr 02 '12 at 16:16

2 Answers2

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From the discussion in the comments two possible solutions arised, here is the first:

Since the group is nilpotent, upper central series terminates. Then for some $n$, $Z_n(G) = G$ and there must be a smallest $i$ such that $N \cap Z_i(G) = \langle e \rangle$ but $N \cap Z_{i+1}(G) \neq \langle e \rangle$. The goal is to show that $N \cap Z_{i+1}(G) \subseteq Z_1(G) = Z(G)$.

I will denote $Z_i(G),Z_{i+1}(G)$ simply by $Z_i, Z_{i+1}$.

We notice that the subgroup $[G,N]$ generated by the elements $\{ gng^{-1}n^{-1}\mid g \in G, n \in N\}$ is contained in the normal subgroup $N$. This is because for every generator $gng^{-1}n^{-1}$ of $[G,N]$, $gng^{-1}n^{-1} = (gng^{-1})n^{-1} \in N$. So in conclusion $[G,N] \subseteq N$.

Further we notice that since $Z(G/Z_i) = Z_{i+1}/Z_i$, then for any $c \in Z_{i+1}, g \in G$ we have that $Z_igc = Z_icg$, which implies $Z_igcg^{-1}c^{-1} = Z_i$ and $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. These are exactly the generators of the group $[G, Z_{i+1}]$, therefore $[G, Z_{i+1}] \subseteq Z_i$

Now we look again at the nontrivial group $N \cap Z_{i+1}$. We established earlier that $[G,N] \subseteq N$ and $[G, Z_{i+1}] \subseteq Z_i$. This gives $[G, N \cap Z_{i+1}] \subseteq [G,N] \cap [G,Z_{i+1}] \subseteq N \cap Z_i = \langle e\rangle$. So actually the subgroup $[G, N \cap Z_{i+1}]$ is trivial, which happens when $N \cap Z_{i+1} \subseteq Z_1 = Z(G)$. This gives that $N \cap Z(G) \neq \langle e \rangle$.

And a perhaps a bit simpler one, relying on an alternative characterisation of nilpotent groups:

By an exercise in my book (Hungerford's Algebra, chapter 2, section 7, exercise 4), a group is nilpotent if and only if the $\gamma_m(G) = \langle e \rangle$ for some $m$, where $\gamma_1(G)=G,\gamma_2(G)=[G,G]$ and $\gamma_i(G)=[\gamma_{i−1}(G),G]$.

Given this, we define a sequence $N_1(G) = N, N_2(G) = [N,G]$ and $N_i(G) = [N_{i-1}(G),G]$, where $N$ is any proper normal subgroup of $G$.

Obviously for any $i$, $N_i(G) \subset N$ by normality of $N$.

It is also clear that $N_1(G) = N \subset G = \gamma_1(G)$. Assume inductively that $N_i(G) \subset \gamma_i(G)$. By definition, $N_{i+1}(G) = [N_{i}(G),G]$ and $\gamma_{i+1}(G)=[\gamma_{i}(G),G]$. It is clear then that $N_{i+1}(G) \subset \gamma_{i+1}(G)$.

But since $G$ is nilpotent, for some $m$, $N_m(G) \subset \gamma_m(G) = \langle e \rangle$ So in particular we get that $[N_{m-1}(G), G] = \langle e \rangle$. This can only be if $N_{m-1}(G) \subset Z(G)$, and since $N_{m-1}(G) \subset N$, we have that $N \cap Z(G) \neq \langle e \rangle$.

Community
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Bartosz Malman
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  • I have a question what happen when G is finite? – Rosa Maria Gtz. Sep 30 '13 at 17:01
  • @Knight Since $p$ groups are nilpotent, this argument applies to $p$-groups. Then any finite nilpotent group is the internal direct sum of its $p$-groups. – mez May 06 '14 at 20:26
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    I think there is a small error in the last part of your argument. How do we know that $N_{m-1}(G)$ is not trivial? You need to insert something to the effect of "This means that there is some $k$ such that $N_k(G)=0$ but $N_{k-1}(G) \neq 0$ (for if there were not we would have that $N=0$)" before your last sentence and replace all m's in it by k's. Note that I am using $0$ to denote the subgroup generated by the identity element. – Nex Oct 14 '15 at 10:48
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Solution:

First we prove that there exists $\{e\}=G_0^*\subset G_1^*\subset ...\subset G_k^*=G$, were $G_{i+1}/G_i=Z(G/G_i)$ and $G_i^*$ is normal subgroup in $G$. To see this we can change groups $\{e\}=G_0\subset G_1\subset ...\subset G_k=G$ to $G_i^{1}=<G_i,Z(G)>,G_0^{1}=\{e\}$, then easy to see that $\{e\}=G_0^{1}\subset G_1^{1}\subset ...\subset G_k^{1}=G$ and $G_1^{1}/G_0^{1}=Z(G)$, like the same define $G_i^2, G_i^3,...$, then $G_i^*=G_i^{k}$.

Now use Induction on $k$, take homomorfism $\pi: G\to G/Z(G)$, then from Induction we get that $\pi(H)\cap \pi(G_2^*)=\pi(H)\cap Z(\pi(G))\not= \{e\}$, so $\exists g\in Z(\pi(G))\not= \{e\}$, $\pi^{-1}(g)= g_2Z(G), g_2\in G_2^*$, $\pi^{-1}(g)\subset HZ(G)$, so $g_2\in HZ(G)$ and $H\cap G_2^*\not= \{e\}$, and we can take $H^*=H\cap G_2^*$, $H^*$ normal subgroup in $G_2$, $G_2/G_1$ is Abelian, $H^*\cap H_1= \{e\}$ so $H^*$ is Abelian.

$\forall g\in G, h^*\in H^*$, $gh^*g^{-1}h^{*-1}\in H^*\cap G'$, $G_2/G_1=Z(G/G_1)$, so $[h^*,g]\in G_1$, but $H\cap G_1=\{e\}$, so $gh^*=hg^*$, $H^*\subset Z(G)=G_1$, so $H\cap Z(G)\not= \{e\}$. done

solver6
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