Some basic properties of nilpotent groups

Prove the following for $G$ an infinite nilpotent group.

1. Let $G$ be a nilpotent group. If $H \leq G$ is a nontrivial normal subgroup, then $H \cap Z(G)$ is nontrivial. In particular, every normal subgroup of prime order is in the center.
2. Let $G$ be a nilpotent group. If $H < G$ is a proper subgroup, then $H < N_G(H)$ is a proper subgroup.

First we prove some lemmas.

Lemma 1: Let $G$ be a group, and let $Z_k(G)$ denote the $k$-th term in the upper central series of $G$. Then $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$. Proof: We proceed by induction on $k$. For the base case, if $k = 0$, we have $Z_0(G/Z(G)) = 1$ $= Z(G)/Z(G)$ $= Z_1(G)/Z(G)$. For the inductive step, suppose $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$ for some $k \geq 0$. We have the following chain of equivalent statements.

 $x Z(G) \in Z_{k+1}(G/Z(G))$ $\Leftrightarrow$ $(xZ(G))Z_k(G/Z(G)) \in Z((G/Z(G))/Z_k(G/Z(G)))$ $\Leftrightarrow$ For all $yZ(G) \in G/Z(G)$, $(xyZ(G))Z_k(G/Z(G)) = (yxZ(G))Z_k(G/Z(G))$ $\Leftrightarrow$ For all $yZ(G) \in G/Z(G)$, $[x,y]Z(G) \in Z_k(G/Z(G))$ $\Leftrightarrow$ For all $y \in G$, $[x,y]Z(G) \in Z_{k+1}(G)/Z(G)$ $\Leftrightarrow$ For all $y \in G$, $[x,y] \in Z_{k+1}(G)$ $\Leftrightarrow$ For all $y \in G$, $xy Z_{k+1}(G) = yx Z_{k+1}(G)$ $\Leftrightarrow$ $x Z_{k+1}(G) \in Z(G/Z_{k+1}(G))$ $\Leftrightarrow$ $x Z_{k+1}(G) \in Z_{k+2}(G)/Z_{k+1}(G)$ $\Leftrightarrow$ $x \in Z_{k+2}(G)$ $\Leftrightarrow$ $xZ(G) \in Z_{k+2}(G)/Z(G)$

Thus $Z_{k+1}(G/Z(G)) = Z_{k+2}(G)/Z(G)$, and the conclusion holds for all $k$. $\square$

Lemma 2: Let $G$ be a group. If $G$ is nilpotent of class $k+1$, then $G/Z(G)$ is nilpotent of class $k$. Proof: $G/Z(G)$ is nilpotent. Using Lemma 1, we have $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$ $= G/Z(G)$, so the nilpotence class of $G/Z(G)$ is at most $k$. Now for $t < k$, we have $Z_t(G/Z(G)) = Z_{t+1}(G)/Z(G)$; since $G$ has nilpotence class $k+1$, $Z_{t+1}(G) \neq G$, and $Z_t(G/Z(G)) \neq G/Z(G)$. Thus the nilpotence class of $G/Z(G)$ is precisely $k$. $\square$

We now move to the main result.

1. We proceed by induction on the nilpotence class of $G$.

For the base case, if $G$ has nilpotence class $k =1$, then $G$ is abelian. Now any nontrivial normal subgroup $H \leq G$ satisfies $H \cap Z(G) = H \neq 1$.

For the inductive step, suppose the conclusion holds for any nilpotent group of nilpotence class $k$. $G$ be a nilpotent group of class $k+1$, and let $H \leq G$ be a nontrivial normal subgroup. Suppose by way of contradiction that $H \cap Z(G) = 1$; now consider the internal direct product $HZ(G) \leq G$. We have $HZ(G)/Z(G) \leq G/Z(G)$; moreover, since $H$ and $Z(G)$ are normal, $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By Lemma 2, $G/Z(G)$ is nilpotent of nilpotence class $k$, so that, by the induction hypothesis, $HZ(G)/Z(G) \cap Z(G/Z(G))$ is nontrivial. Let $xzZ(G) \in HZ(G)/Z(G) \cap Z(G/Z(G))$, where $x \in H$ and $z \in Z(G)$. In fact, we have $xZ(G) \in Z(G/Z(G))$. Let $g \in G$ be arbitrary. Since $H$ is normal in $G$, $g^{-1}hg \in H$. Now $g^{-1}hgZ(G) = hZ(G)$ since $hZ(G)$ is central in $G/Z(G)$, so that $h^{-1}g^{-1}hg \in Z(G)$. Recall, however, that $H \cap Z(G) = 1$, so that $h^{-1}g^{-1}hg = 1$, and we have $gh = hg$. Thus $h \in Z(G)$, a contradiction. So $H \cap Z(G) \neq 1$.

2. We proceed again by induction on the nilpotence class of $G$.

For the base case, if $G$ has nilpotence class $k = 1$, then $G$ is abelian. Thus $N_G(H) = G$ for all subgroups $H$, and if $H < G$ is proper, then $H < N_G(H)$ is proper.

For the inductive step, suppose every nilpotent group of nilpotence class $k \geq 1$ has the desired property. Let $G$ be a nilpotent group of nilpotence class $k+1$, and let $H < G$ be a proper subgroup. Now $G/Z(G)$ is nilpotent of class $k$. Suppose $Z(G) \not\leq H$; then, since $Z(G) \leq N_G(H)$, $H \leq \langle H, Z(G) \rangle$ is proper, and $H \leq N_G(H)$ is proper. If $Z(G) \leq H$, then $H/Z(G) \leq G/Z(G)$ is proper. By the induction hypothesis, $H/Z(G) \leq N_{G/Z(G)}(H/Z(G)) = N_G(H)/Z(G)$ is proper, so that $H \leq N_G(H)$ is proper.