At most n elements of order dividing n implies every finite subgroup is cyclic
This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., group with at most n elements of order dividing n) must also satisfy the second group property (i.e., group in which every finite subgroup is cyclic)
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Statement with symbols
Suppose is a group, with identity element , with the property that for any , there are at most elements satisfying . Then, every finite subgroup of is cyclic.
- Multiplicative group of a field implies every finite subgroup is cyclic
- Multiplicative group of a prime field is cyclic
- Unique subgroup for every divisor of order iff cyclic: A finite group of order is cyclic if and only if it has a unique subgroup of order for every positive integer dividing .
- Number of nth roots is a multiple of n
- Number of nth roots of any conjugacy class is a multiple of n
- Frobenius conjecture on nth roots
- Elements of multiplicative group equal generators of additive group
- Finite group is disjoint union of sets of cyclic elements for cyclic subgroups
- Natural number equals sum of Euler-phi function values at divisors
- Order of element divides order of group
Given: A group with the property that for every , there are at most solutions to .
To prove: Every finite subgroup of is cyclic.
Proof: Let denote the Euler-phi function of : in other words, is the number of elements that generate a given cyclic group of order (see fact (1) for more on this set of elements).
By fact (3), we then have, for any natural number :
Let be the set of all elements such that and denote the set of all elements of order exactly . Then, since implies that the order of divides , we get:
Thus, if denote the cardinalities of respectively, we get:
Next, we observe that for any : If , we are done. Otherwise, there exists a of order exactly . The cyclic subgroup in is a cyclic group of order . This has elements satisfying . Since there are at most elements in satisfying , we conclude that . In this case, it is clear that is precisely the set of generators of , which has size . Thus, .
We now prove the claim by induction on order.
- Base case: The result is clearly true for a subgroup of order .
- Induction step: Suppose is a finite subgroup of of order . Note that every element of has order dividing (fact (4)), so . But since there are at most elements of order , we in fact have with . On the other hand, we have for all , so for all divisors of . Thus, we get:
Equality must hold throughout, forcing for all . In particular, , so there exists an element of order exactly . Further, since , this element must be in , and we thus get that is the cyclic group of order on this element.