Basic properties of finite nilpotent groups

Let G be a finite nilpotent group.

  1. If H is a nontrivial normal subgroup of G, then H intersects the center of G nontrivially. In particular, every normal subgroup of prime order is central.
  2. If H < G is a proper subgroup then H is properly contained in N_G(H).

First we prove some lemmas.

Lemma 1: Let G be a group, with A_1,A_2,B_1,B_2 \subseteq G. If A_1 \subseteq A_2 and B_1 \subseteq B_2, then [A_1,B_1] \leq [A_2,B_2]. Proof: [A_1,B_1] is generated by [a,b] such that a \in A_1 and b \in B_1. Each of these generators is in [A_2,B_2]. \square

Lemma 2: Let G be a group. If H \leq G, then H^k \leq G^k for all k, where G^k denotes the k-th term in the lower central series of G. Proof: We proceed by induction on k. For the base case, if k=0 then H^0 \leq G^0. For the inductive step, suppose H^k \leq G^k for some k \geq 0. Then by Lemma 1, H^{k+1} = [H,H^k] \leq [G,G^k] = G^{k+1}. \square

Lemma 3: If G is a nilpotent group of nilpotence class k and H \leq G, then H is nilpotent with nilpotence class at most k. Proof: By Theorem 8 in the text, G^k = 1. By Lemma 2, H^k = 1, so that H is nilpotent and of nilpotence class at most k. \square

Lemma 4: Let \varphi : G \rightarrow H be a group homomorphism and let S \subseteq G. Then \varphi[\langle S \rangle] = \langle \varphi[S] \rangle. Proof: We have x \in \varphi[\langle S \rangle] if and only if x = \varphi(s_1^{a_1} \cdots s_k^{a_k}) for some s_i \in S if and only if x = \varphi(s_1)^{a_1} \cdots \varphi(s_k)^{a_k} for some s_i \in S if and only if x \in \langle \varphi[S] \rangle. \square

Lemma 5: Let \varphi : G \rightarrow H be a surjective group homomorphism. Then H^k \leq \varphi[G^k] for all k, where G^k denotes the k-th term in the lower central series of G. Proof: We proceed by induction on k. For the base case, if k = 0 then H^0 = H = \varphi[G] = \varphi[G^0]. For the inductive step, suppose H^k \leq \varphi[G^k]. Then we have H^{k+1} = [H,H^k] \leq [\varphi[G],\varphi[G^k]] \leq \varphi[[G,G^k]] \leq \varphi[G^{k+1}]. \square

Lemma 6: Let G be a nilpotent group of nilpotence class k. If H \leq G is normal, then G/H is nilpotent of nilpotence class at most k. Proof: By Lemma 5, we have (G/H)^k \leq \varphi[G^k] = 1. \square

Now to the main results; first we show part 2. [With hints from MathReference.]

(2) Let G be a finite nilpotent group such that Z(G) \neq 1. If H \leq G is a nontrivial normal subgroup of G, then H \cap Z(G) is nontrivial. In particular, every normal subgroup of order p is contained in the center.

We proceed by induction on the breadth of G. (Recall: the breadth of a finite group is the number of prime factors dividing its order, including multiplicity.)

For the base case, if G has breadth 1, then G \cong Z_p for some prime p and thus G is simple and abelian. Thus H = G, and H \cap Z(G) \neq 1.

For the inductive step, suppose that the conclusion holds for every finite nilpotent group of breadth at most k \geq 1, and let G be a finite nilpotent group of breadth k+1. Let H \leq G be normal and suppose H \cap Z(G) = 1. Now H \cdot Z(G) \cong H \times Z(G) \leq G is an internal direct product. Moreover, since H and Z(G) are normal in G, H \cdot Z(G) \leq G is normal. Now by the Lattice Isomorphism Theorem, HZ(G)/Z(G) \leq G/Z(G) is normal. By Lemma 6, G/Z(G) is nilpotent, has breadth at most k, and HZ(G)/Z(G) \leq G/Z(G) is normal. By the inductive hypothesis, there exists an element xZ(G) \in Z(G/Z(G)) such that xZ(G) \in HZ(G)/Z(G). We can write x = hz, where h \in H and z \in Z(G), so that in fact hZ(G) \in Z(G/Z(G)). Now let g \in G be arbitrary; since H is normal in G, ghg^{-1} \in H. Now (ghg^{-1})Z(G) = (gZ(G))(hZ(G))(g^{-1}Z(G)) = hZ(G) since hZ(G) is in the center of G/Z(G), so that ghg^{-1}Z(G) = hZ(G). Write ghg^{-1} = hk, where k \in Z(G). Now ghg^{-1}h^{-1} = k \in Z(G), and since H \cap Z(G) = 1, we have ghg^{-1}h^{-1} = 1, thus gh = hg. So h \in H is in the center of G and h \neq 1, a contradiction. Thus H \cap Z(G) is nontrivial. \square

Next, we prove some more lemmas.

Lemma 7: Let G be a group, K, H_1, H_2 \leq G be subgroups with K \leq G normal, K \leq H_1, and K \leq H_2. If H_1/K = H_2/K, then H_1 = H_2. Proof: Let x \in H_1. Then xK \in H_1/K = H_2/K, so that xK = yK. In particular, x \in yK \leq H_2. The other direction is similar. \square

Lemma 8: Let G be a group, H \leq G a normal subgroup, and K a subgroup with H \leq K \leq G. Then N_{G/H}(K/H) = N_G(K)/H. Proof: Note that xH \in N_{G/H}(K/H) if and only if (xH)(K/H)(x^{-1}H) = K/H, if and only if (xKx^{-1})/H = K/H, if and only if (by Lemma 7) xKx^{-1} = K, if and only if x \in N_G(K), if and only if xH \in N_G(K)/H. \square

Now we show part 4.

(4) Let G be a finite nilpotent group. If H < G is a proper subgroup, then H < N_G(H) is proper.

Proof: We proceed again by induction on the breadth of G.

For the base case, if G has breadth 1, then G \cong Z_p is abelian. Then if H \leq G is proper, H \leq N_G(H) = G is proper.

For the inductive step, suppose the conclusion holds for all finite nilpotent groups of breadth at most k. Let G be a finite nilpotent group of breadth k+1 and let H \leq G be a proper subgroup. Now Z(G) \leq N_G(H). If Z(G) \not\leq H, then H \leq \langle H, Z(G) \rangle is proper, so that H \leq N_G(H) is proper. If Z(G) \leq H, then H/Z(G) \leq G/Z(G) is proper. Moreover, G/Z(G) is a finite nilpotent group of breadth at most k, so that H/Z(G) \leq N_{G/Z(G)}(H/Z(G)) is proper. By Lemma 8, H/Z(G) \leq N_G(H)/Z(G) is proper, so that H \leq N_G(H) is proper. \square

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