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, a homomorph-containing subgroup and a variety-containing subgroup.
- Number of nth roots is a multiple of n
- Number of nth roots of any conjugacy class is a multiple of n
- Exactly n elements of order dividing n implies every finite subgroup is cyclic
- Frobenius conjecture on nth roots: The conjecture states that the assumption of solvability can be dropped.
- Minimal normal implies elementary abelian in finite solvable
- Lagrange's theorem: We in particular are interested in the version which states that if is a subgroup of , , where is the quotient set. When is a normal subgroup, is the quotient group.
- Number of nth roots is a multiple of n
- Solvability is quotient-closed
- Hall subgroups exist in finite solvable
This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).
We do the proof by induction. Specifically, we assume that the statement holds true for all finite solvable groups of strictly smaller orders than .
PROOF OF INDUCTIVE STEP:
Given: A finite solvable group of order . 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
|No.||Assertion/construction||Facts used||Given data/assumptions used||Previous steps used||Explanation|
|1||The order of every element in is either or||is elementary abelian of order|
|2||divides , comprises all elements whose power is the identity||Step (1)||[SHOW MORE]|
|3||Let and where are relatively prime to . Then, and .||divides the order of||[SHOW MORE]|
|4||has order||Fact (2)||has order||Step (3)||[SHOW MORE]|
|5||Let . Then, divides the order of .||Steps (3), (4)||[SHOW MORE]|
|6||There exists some natural number such that has elements of order dividing||Fact (3)||Step (5)||[SHOW MORE]|
|7||Let be the quotient map. The set of elements such that is the identity element of has size||has order||Step (6)||[SHOW MORE]|
|8||If is the identity element in , then is the identity element in .||Step (1)||[SHOW MORE]|
|9||divides .||divides||Steps (3), (5)||[SHOW MORE]|
|12||and is the identity element iff . In other words, .||Steps (5), (11)||[SHOW MORE]|
|13||The set of elements such that is the identity element forms a subgroup. In particular, is a subgroup||Fact (4)||inductive hypothesis, is solvable||Step (12)||[SHOW MORE]|
|14||is a subgroup||Step (13)||[SHOW MORE]|
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.
Proof that the subgroup is normal, characteristic and fully characteristic
This follows from the fact that if is a homomorphism from to any subgroup of , then . Hence, if , so is .