A nilpotent group can be thought of a group that is only finitely removed from an abelian group. Specifically, it is a group such that is the trivial group, for some integer , where is the th term of the lower central series of . The least integer satisfying this condition is called the nilpotency class of . Using transfinite recursion, the notion of nilpotency class can be extended to any ordinal.
Characterization and Properties of Nilpotent Groups
Theorem 1. Let be a group, and let be a positive integer. Then the following three statements are equivalent:
- The group has nilpotency class at most ;
- There exists a sequence of subgroups of such that , for all integers .
- For every subgroup of , there exist subgroups , such that , , and is a normal subgroup of such that is commutative, for all integers .
- The group has a subgroup in the center of such that has nilpotency class at most .
Proof. To show that (1) implies (2), we may take .
To show that (2) implies (1), we note that it follows from induction that ; hence .
Now, we show that (1) implies (3). Set ; we claim that this suffices. We wish first to show that normalizes . Since evidently normalizes , it suffices to show that does; to this end, let be an element of and an element of . Then Thus normalizes . To prove that is commutative, we note that is commmutative, and that the canonical homomorphism from to is surjective; thus is commutative.
To show that (3) implies (1), we may take .
To show that (1) implies (4), we may take .
Finally, we show that (4) implies (1). Let be the canonical homomorphism of onto . Then . In particular, . Hence is a subset of , so it lies in the center of , and ; thus the nilpotency class of is at most , as desired.
Corollary 2. Let be a nilpotent group; let be a subgroup of . If is its own normalizer, then .
Proof. Suppose ; then there is a greatest integer for which . Then normalizes .
Corollary 3. Let be a nilpotent group; let be a proper subgroup of . Then there exists a proper normal subgroup of such that and is abelian.
Proof. In the notation of the theorem, let be the least integer such that . Then set .
Corollary 4. Let be a nilpotent group; let be a subgroup of . If , then .
Proof. Suppose that . Then let be the normal subgroup of containing as described in Corollary 3. Then , so a contradiction.
Corollary 5. Let be a group, let be a nilpotent group, and let be a group homomorphism for which the homomorphism derived from passing to quotients is surjective. Then is surjective.
Proof. Let be the image of and apply Corollary 3.
Proposition. Let be a group of nilpotency class at most , and let be a normal subgroup of . Then there exists a sequence of subgroups of such that , , , and , for all integers .
Proof. Let . Then and since is a normal subgroup.
Corollary 6. Let be a nilpotent group; let be a normal subgroup of , and let be the center of . If is not trivial, then is not trivial.
Proof. In the proposition's notation, let be the greatest integer such that . The , so is a nontrivial subgroup that lies in the center of and in .
Corollary 7. Let be a nilpotent group, let be a group, and let be a homomorphism of into . If the restriction of to the center of is injective, then so is .
Proof. We proceed by contrapositive. Suppose that is not injective; then the kernel of is nontrivial, so by the previous corollary, the intersection of and the center of is nontrivial, so the restriction of to the center of is not injective.
Finite Nilpotent Groups
Theorem 8. Let be a finite group. Then the following conditions are equivalent.
Proof. Since every -group is nilpotent, condition (2) implies condition (1).
Now we show that (1) implies (3). Let be a Sylow -subgroup of , and let be its normalizer. Then is its own normalizer. Then from Corollary 2, , i.e., is normal in .
Finally, we show that (3) implies (2). Suppose condition (3) holds for . For any prime dividing the order of , let denote the Sylow -subgroup of . Let and be distinct primes dividing the order of . Then , since the order of any element in both of these groups must divide a power of and a power of . Since and are both normal, it follows that for any , , the commutator is an element both of and . It follows that the canonical mapping of is a homomorphism, and is in its image, for every prime . Now, the order subgroup generated by the must be divisible by every power of a prime that divides , but it must also divide ; hence it is equal to . It follows that the order of the image of is equal to the order of ; since is finite, this implies that is surjective. Since and have the same size, is also injective, and hence an isomorphism.