At most n elements of order dividing n implies every finite subgroup is cyclic

Property-theoretic statement
The group property of being a group with at most n elements of order dividing n is stronger than the group property of being a group in which every finite subgroup is cyclic.

Statement with symbols
Suppose $$G$$ is a group, with identity element $$e$$, with the property that for any $$n$$, there are at most $$n$$ elements $$x \in G$$ satisfying $$x^n = e$$. Then, every finite subgroup of $$G$$ is cyclic.

Applications

 * 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 $$n$$ is cyclic if and only if it has a unique subgroup of order $$d$$ for every positive integer $$d$$ dividing $$n$$.

Other related facts

 * 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

Facts used

 * 1) uses::Elements of multiplicative group equal generators of additive group
 * 2) uses::Finite group is disjoint union of sets of cyclic elements for cyclic subgroups
 * 3) uses::Natural number equals sum of Euler-phi function values at divisors
 * 4) uses::Order of element divides order of group

Proof
Given: A group $$G$$ with the property that for every $$n$$, there are at most $$n$$ solutions to $$x^n = 1$$.

To prove: Every finite subgroup of $$G$$ is cyclic.

Proof: Let $$\varphi(d)$$ denote the Euler-phi function of $$d$$: in other words, $$\varphi(d)$$ is the number of elements that generate a given cyclic group of order $$d$$ (see fact (1) for more on this set of elements).

By fact (3), we then have, for any natural number $$n$$:

$$n = \sum_{d|n} \varphi(d)$$

Let $$A(d)$$ be the set of all elements $$x \in G$$ such that $$x^d = 1$$ and $$B(d)$$ denote the set of all elements of order exactly $$d$$. Then, since $$x^n = 1$$ implies that the order of $$x$$ divides $$n$$, we get:

$$A(n) = \bigsqcup_{d|n} B(d)$$.

Thus, if $$\alpha(n), \beta(n)$$ denote the cardinalities of $$A(n), B(n)$$ respectively, we get:

$$\alpha(n) = \sum_{d|n} \beta(d)$$.

Next, we observe that $$\beta(d) \in \{ 0, \varphi(d) \}$$ for any $$d$$: If $$\beta(d) = 0$$, we are done. Otherwise, there exists a $$g \in G$$ of order exactly $$d$$. The cyclic subgroup $$\langle g \rangle$$ in G is a cyclic group of order $$d$$. This has $$d$$ elements satisfying $$x^d = 1$$. Since there are at most $$d$$ elements in $$G$$ satisfying $$x^d = 1$$, we conclude that $$A(d) = \langle g \rangle$$. In this case, it is clear that $$B(d)$$ is precisely the set of generators of $$\langle g \rangle$$, which has size $$\varphi(d)$$. Thus, $$\beta(d) = \varphi(d)$$.

We now prove the claim by induction on order.


 * Base case: The result is clearly true for a subgroup of order $$1$$.
 * Induction step: Suppose $$H$$ is a finite subgroup of $$G$$ of order $$n$$. Note that every element of $$H$$ has order dividing $$n$$ (fact (4)), so $$\alpha(n) \ge n$$. But since there are at most $$n$$ elements of order $$n$$, we in fact have $$\alpha(n) = n$$ with $$H = A(n)$$. On the other hand, we have $$\beta(d) \in \{ 0, \varphi(d) \}$$ for all $$d$$, so $$\beta(d) \le \varphi(d)$$ for all divisors $$d$$ of $$n$$. Thus, we get:

$$n = \alpha(n) = \sum_{d|n} \beta(d) \le \sum_{d|n} \varphi(d) = n$$.

Equality must hold throughout, forcing $$\beta(d) = \varphi(d)$$ for all $$d | n$$. In particular, $$\beta(n) \ne 0$$, so there exists an element of order exactly $$n$$. Further, since $$H = A(n)$$, this element must be in $$H$$, and we thus get that $$H$$ is the cyclic group of order $$n$$ on this element.