Let be a finite nilpotent group.

- If is a nontrivial normal subgroup of , then intersects the center of nontrivially. In particular, every normal subgroup of prime order is central.
- If is a proper subgroup then is properly contained in .

First we prove some lemmas.

Lemma 1: Let be a group, with . If and , then . Proof: is generated by such that and . Each of these generators is in .

Lemma 2: Let be a group. If , then for all , where denotes the -th term in the lower central series of . Proof: We proceed by induction on . For the base case, if then . For the inductive step, suppose for some . Then by Lemma 1, .

Lemma 3: If is a nilpotent group of nilpotence class and , then is nilpotent with nilpotence class at most . Proof: By Theorem 8 in the text, . By Lemma 2, , so that is nilpotent and of nilpotence class at most .

Lemma 4: Let be a group homomorphism and let . Then . Proof: We have if and only if for some if and only if for some if and only if .

Lemma 5: Let be a surjective group homomorphism. Then for all , where denotes the -th term in the lower central series of . Proof: We proceed by induction on . For the base case, if then . For the inductive step, suppose . Then we have .

Lemma 6: Let be a nilpotent group of nilpotence class . If is normal, then is nilpotent of nilpotence class at most . Proof: By Lemma 5, we have .

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

(2) Let be a finite nilpotent group such that . If is a nontrivial normal subgroup of , then is nontrivial. In particular, every normal subgroup of order is contained in the center.

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

For the base case, if has breadth 1, then for some prime and thus is simple and abelian. Thus , and .

For the inductive step, suppose that the conclusion holds for every finite nilpotent group of breadth at most , and let be a finite nilpotent group of breadth . Let be normal and suppose . Now is an internal direct product. Moreover, since and are normal in , is normal. Now by the Lattice Isomorphism Theorem, is normal. By Lemma 6, is nilpotent, has breadth at most , and is normal. By the inductive hypothesis, there exists an element such that . We can write , where and , so that in fact . Now let be arbitrary; since is normal in , . Now since is in the center of , so that . Write , where . Now , and since , we have , thus . So is in the center of and , a contradiction. Thus is nontrivial.

Next, we prove some more lemmas.

Lemma 7: Let be a group, be subgroups with normal, , and . If , then . Proof: Let . Then , so that . In particular, . The other direction is similar.

Lemma 8: Let be a group, a normal subgroup, and a subgroup with . Then . Proof: Note that if and only if , if and only if , if and only if (by Lemma 7) , if and only if , if and only if .

Now we show part 4.

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

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

For the base case, if has breadth 1, then is abelian. Then if is proper, is proper.

For the inductive step, suppose the conclusion holds for all finite nilpotent groups of breadth at most . Let be a finite nilpotent group of breadth and let be a proper subgroup. Now . If , then is proper, so that is proper. If , then is proper. Moreover, is a finite nilpotent group of breadth at most , so that is proper. By Lemma 8, is proper, so that is proper.