Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup
Suppose is a Finite solvable group (?), and is a natural number dividing the order of . If there are exactly elements of whose power is the identity element, then these elements form a subgroup. This subgroup is clearly a normal subgroup. In fact, it is a characteristic subgroup, and even better, is a fully characteristic subgroup.
- Minimal normal implies elementary abelian in finite solvable
- Number of nth roots is a multiple of n
- Solvability is quotient-closed
Given: A group . A natural number . is the set of roots of unity in (i.e., the set of elements of order dividing , and has elements.
To prove: is a subgroup.
Proof: By fact (1), has a nontrivial elementary abelian normal subgroup of order for some prime power dividing the order of . We split the proof into two cases, based on whether divides or not.
Case that divides
In this case, . Let and where are relatively prime to . Clearly, and .
- Let . Then, the order of is divisible by : This is clear from the fact that the order of equals , and since and , this follows.
- divides : This follows essentially from the fact that exponent of in is strictly less than that in .
- There exists some natural number such that has elements of order dividing : This follows from fact (2).
- Let be the quotient map. The set of elements such that has size : This follows from step (3).
- If , we have : This follows from the fact that the order of any element in is either or .
- If , we have : This follows from the previous step and step (2), which says that .
- , and : By steps (3) and (5), there are at least elements for which . Thus, . Simplifying, we obtain that . This forces and . Further, since equality holds even under these assumptions, we conclude that the solutions to exhaust all solutions to .
- The set of elements such that forms a subgroup. In particular, is a subgroup: By the previous step, , so there are exactly solutions in to . By fact (3), is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.
- is a subgroup: It is the inverse image in of a subgroup of .
Case that does not divide
- divides the order of : This follows because divides the order of and is relatively prime to .
- The number of elements such that is for some natural number : This follows from fact (2).
- For every coset of in such that , there exists an element such that : Start with some element in the coset. Let . By definition, . Now, find an integer such that (this can be done since and are relatively prime). Let and let . Note that since is a power of , which in turn is a power of , commutes with . Thus, if , then since is a multiple of , and . Further, , so . Thus, we have found an element of whose order divides .
- There are at least solutions to in : This follows from steps (2) and (3).
- , and there are exactly solutions to in : This follows from the previous step and the assumption that there are exactly solutions.
- The set of solutions to in forms a subgroup, say : By fact (3), is solvable, so we can apply the induction hypothesis to the conclusion of the previous step.
- The inverse image, say of this subgroup of in is a group of order , where is relatively prime to .
- has a -complement, i.e., a Hall subgroup of order : This follows from fact (4).
- This Hall subgroup of order is precisely equal to the set of solutions to : All elements in the Hall subgroup of order are solutions to , and we know that there are exactly solutions. Hence, these must be the only solutions.