Prove the following for an infinite nilpotent group.
- Let be a nilpotent group. If is a nontrivial normal subgroup, then is nontrivial. In particular, every normal subgroup of prime order is in the center.
- Let be a nilpotent group. If is a proper subgroup, then is a proper subgroup.
First we prove some lemmas.
Lemma 1: Let be a group, and let denote the -th term in the upper central series of . Then . Proof: We proceed by induction on . For the base case, if , we have . For the inductive step, suppose for some . We have the following chain of equivalent statements.
For all , | |
For all , | |
For all , | |
For all , | |
For all , | |
Thus , and the conclusion holds for all .
Lemma 2: Let be a group. If is nilpotent of class , then is nilpotent of class . Proof: is nilpotent. Using Lemma 1, we have , so the nilpotence class of is at most . Now for , we have ; since has nilpotence class , , and . Thus the nilpotence class of is precisely .
We now move to the main result.
- We proceed by induction on the nilpotence class of .
For the base case, if has nilpotence class , then is abelian. Now any nontrivial normal subgroup satisfies .
For the inductive step, suppose the conclusion holds for any nilpotent group of nilpotence class . be a nilpotent group of class , and let be a nontrivial normal subgroup. Suppose by way of contradiction that ; now consider the internal direct product . We have ; moreover, since and are normal, is normal. By Lemma 2, is nilpotent of nilpotence class , so that, by the induction hypothesis, is nontrivial. Let , where and . In fact, we have . Let be arbitrary. Since is normal in , . Now since is central in , so that . Recall, however, that , so that , and we have . Thus , a contradiction. So .
- We proceed again by induction on the nilpotence class of .
For the base case, if has nilpotence class , then is abelian. Thus for all subgroups , and if is proper, then is proper.
For the inductive step, suppose every nilpotent group of nilpotence class has the desired property. Let be a nilpotent group of nilpotence class , and let be a proper subgroup. Now is nilpotent of class . Suppose ; then, since , is proper, and is proper. If , then is proper. By the induction hypothesis, is proper, so that is proper.