Problem 6.1.2 (Properties of nilpotent groups)

We will use the following Lemma:

Lemma. If $G$ is nilpotent, then $G∕Z(G)$ is nilpotent.

We prove

(a)

Theorem A. Let $G$ be a finite nilpotent group. If $N$ is a nontrivial normal subgroup of $G$ then $N$ intersects the center non-trivially: $$N\cap Z(G)\ne \{1\}.$$

First proof (my proof).

Proof. For the sake of contradiction, consider a counterexample $G$ of this Theorem of minimal order, so that there is a nontrivial subgroup of $G$ such that

$$\begin{array}{lll}\hfill N\cap Z(G)=\{1\}.& & \hfill \text{(3)}\end{array}$$

Since $G$ is nilpotent, the center $Z(G)=Z_1(G)\ne \{1\}$. Let $\bar{G}=G∕Z(G)$. Then $|\bar{G}|<|G|$, and $G∕Z(G)$ is a finite nilpotent group by the Lemma, so by minimality, $\bar{G}$ is such that every nontrivial normal subgroup of $G$ intersects the center non-trivially.

Since $N$ and $Z(G)$ are normal subgroups of $G$, then $\mathit{𝑁𝑍}(G)$ is a normal subgroup of $G$ containing $Z(G)$.

If $\mathit{𝑁𝑍}(G)=Z(G)$, then $N\subseteq Z(G)$, so $N\cap Z(G)=N\ne \{1\}$, which contradicts (3). Therefore $\mathit{𝑁𝑍}(G)\ne Z(G)$, so $(\mathit{𝑁𝑍}(G))∕Z(G)$ is a nontrivial normal subgroup of $\bar{G}$. This shows that

$$(\mathit{𝑁𝑍}(G))∕Z(G)\cap Z(G∕Z(G))\ne \{1\}.$$

Hence there is some $n\in N$ such that $\mathit{𝑛𝑍}(G)\ne Z(G)$ is in the center of $\bar{G}$. Therefore, for every $g\in G$,

$$\mathit{𝑛𝑍}(G)\mathit{𝑔𝑍}(G)=\mathit{𝑔𝑍}(G)\mathit{𝑛𝑍}(G),$$

so $n^{-1}{g}^{-1}\mathit{𝑛𝑔𝑍}(G)=Z(G)$, thus

$$[n,g]=n^{-1}{g}^{-1}\mathit{𝑛𝑔}\in Z(G).$$

But $N⊴G$, thus $n^{-1}(g^{-1}\mathit{𝑛𝑔})\in N$, so

$$[n,g]\in N\cap Z(G).$$

Using $N\cap Z(G)=\{1\}$, we obtain $[n,g]=1$, so $\mathit{𝑛𝑔}=\mathit{𝑔𝑛}$ for all $g\in G$. This means $n\in G\cap Z(G)$, where $n\ne 1$ since $\mathit{𝑛𝑍}(G)\ne Z(G)$, so $G\cap Z(G)\ne \{1\}$. This is a contradiction, which proves Theorem A. □

Second proof(Bartosz Malman)

(Source: Bartosz Malman (https://math.stackexchange.com/users/14660/bartosz-malman), Infinite nilpotent group, any normal subgroup intersects the center nontrivially, URL (version: 2012-04-02): https://math.stackexchange.com/q/127001)

This proof remains true if $G$ is infinite.

Proof. Let $G$ be a finite nilpotent group, and let $N$ be a nontrivial normal subgroup of $G$. Let us denote $Z_i=Z_i(G)$.

$\bullet$ There is some index $c$ such that $Z_c=G$, and $Z_0=\{1\}$, then $N\cap Z_0=\{1\}$ and $N\cap Z_c=N\ne \{1\}$. Hence there is a largest integer $i$ ( $0\le i<c$) such that $N\cap Z_i=\{1\}$, so that

$$N\cap Z_i=\{1\}\text{and}N\cap Z_{i+1}\ne \{1\}.$$

We want to prove $N\cap Z_{i+1}\subseteq Z(G)$.

$\bullet$ The subgroup $[N,G]$ is generated by the elements $[n,g]=n^{-1}{g}^{-1}\mathit{𝑛𝑔}$, where $n\in N$ and $g\in G$. Since $N⊴G$, then $[n,g]=n^{-1}(g^{-1}\mathit{𝑛𝑔})\in N$. Therefore

$$\begin{array}{lll}\hfill [N,G]\subseteq N.& & \hfill \text{(4)}\end{array}$$

$\bullet$ By definition of the upper central series of $G$, $Z_{i+1}∕Z_i=Z(G∕Z_i)$. Therefore, for any $c\in Z_{i+1}$ and for any $g\in G$,

$$c{Z}_{i}g{Z}_i=g{Z}_{i}c{Z}_i,$$

thus $c^{-1}{g}^{-1}\mathit{𝑐𝑔}{Z}_i=Z_i$, so $[c,g]=c^{-1}{g}^{-1}\mathit{𝑐𝑔}\in Z_i$. This proves

$$\begin{array}{lll}\hfill [Z_{i+1},G]\subseteq Z_i.& & \hfill \text{(5)}\end{array}$$

$\bullet$ By (4) and (5), we obtain

$$[N\cap Z_{i+1},G]\subseteq [N,G]\cap [Z_{i+1},G]\subseteq N\cap Z_i=\{1\}.$$

Therefore $[N\cap Z_{i+1},G]=\{1\}$. This means that the elements of $N\cap Z_{i+1}$ commute with every element of $G$, so

$$N\cap Z_{i+1}\subseteq Z(G).$$

$\bullet$ Hence $N\cap Z(G)\supseteq N\cap (N\cap Z_{i+1})=N\cap Z_{i+1}\ne \{1\}$, so $N\cap Z(G)\ne \{1\}$.

If $N$ is a nontrivial normal subgroup of $G$ then $N$ intersects the center non-trivially □

(b)

Now we prove

Theorem B. Let $G$ be a finite nilpotent group. If $H<G$ then $H<N_G(H)$ (i.e., every proper subgroup of $G$ is a proper subgroup of its normalizer in $P$).

Proof. As in the text p.189, we prove Theorem B by induction on $|G|$. If $|G|=1$, then $G$ has no proper subgroup, so the proposition is vacuously true.

Suppose that Theorem B is true for every nilpotent group of order less that $n$, and let $G$ be a nilpotent group of order $n>1$. Then $Z(G)=Z_1\ne \{1\}$, and $G∕Z(G)$ is nilpotent by the Lemma, so we may apply the induction hypothesis to $G∕Z(G)$.

Let $H$ be a proper subgroup of $G$. Then $Z(G)\ne \{1\}$ normalizes $H$. If $Z(G)$ is not contained in $H$, then $H$ is properly contained in $⟨H,Z(G)⟩$, and the latter subgroup is contained in $N_G(H)$ so $H<N_G(H)$.

We may therefore assume $Z(P)\le H$. Use bar notation to denote passage to the quotient $G∕Z(G)$. By the induction hypothesis, $\bar{H}$ is properly contained in $N_{\bar{G}}(\bar{H})$:

$$\begin{array}{lll}\hfill \bar{H}<N_{\bar{G}}(\bar{H}).& & \hfill \text{(6)}\end{array}$$

We check $N_{\bar{G}}(\bar{H})=\overline{N_G(H)}$, so that $N_G(H)$ is the complete preimage in $G$ of $N_{\bar{G}}(\bar{H})$: by the Lattice Isomorphism Theorem, $\bar{H}\le \bar{K}$ if and only if $H\le K$ for all subgroups $H,K$ containing $Z(G)$, so for all $g\in G$, $\bar{g}=\mathit{𝑔𝑍}(G)\in \bar{G}$, and

$$\begin{array}{llllll}\hfill \bar{g}\in N_{\bar{G}}(\bar{H})& ⟺\bar{g}\bar{H}{\bar{g}}^{-1}\subseteq \bar{H}& \hfill & & \hfill & \\ \hfill & ⟺\overline{\mathit{𝑔𝐻}{g}^{-1}}\subseteq \bar{H}& \hfill (\text{where }Z(G)=\mathit{𝑔𝑍}(G)g^{-1}\subseteq \mathit{𝑔𝐻}{g}^{-1})& & \hfill & & \hfill \\ \hfill & ⟺\mathit{𝑔𝐻}{g}^{-1}\subseteq H& \hfill & & \hfill & \\ \hfill & ⟺g\in N_G(H)& \hfill & & \hfill & \\ \hfill & ⟺\bar{g}\in \overline{N_G(H)},& \hfill & & \hfill & \end{array}$$

so

$$N_{\bar{G}}(\bar{H})=\overline{N_G(H)}.$$

Then (6) becomes

$$\bar{H}<\overline{N_G(H)},$$

therefore

$$H<N_G(H)$$

(If $H=N_G(H)$, then $\bar{H}=\overline{N_G(H)}$).

The induction is done, which proves Theorem B. □