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.

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