# Infinite nilpotent group, any normal subgroup intersects the center nontrivially

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?

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Consider $[N,G,G,\ldots]$. –  user641 Apr 1 '12 at 19:47
This is called "Normality Large". There is a proof of this on Groupprops: groupprops.subwiki.org/wiki/… –  Alex Youcis Apr 1 '12 at 19:48
@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 1 '12 at 20:42
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 1 '12 at 22:13
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. –  user641 Apr 1 '12 at 22:32

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 $N \cap Z(G) \neq \langle e \rangle$.

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I have a question what happen when G is finite? –  Knight 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 6 at 20:26

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

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