Let \(G\) be a group of order \(n\). Suppose that for every divisor \(d\) of \(n\), \(G\) contains at most one subgroup of order \(d\). Show that \(G\) is cyclic.
Let \(F\) be a field. Show that every finite subgroup of the group of units \(F^\times\) is cyclic.
Solution
Part 1.
1. For each divisor \(d\) of \(n\), let \(a_d\) be the number of elements of \(G\) of order exactly \(d\).
This is a definition.
2. If \(G\) has an element of order \(d\), then it has a cyclic subgroup of order \(d\), which (by hypothesis) is the unique subgroup of order \(d\).
The subgroup generated by an element of order \(d\) has order \(d\).
3. A cyclic group of order \(d\) has exactly \(\varphi(d)\) elements of order \(d\).
If \(g\) generates a cyclic group of order \(d\), then \(g^k\) has order \(d/\gcd(k,d)\), which equals \(d\) exactly when \(\gcd(k,d)=1\). There are \(\varphi(d)\) such \(k\) with \(0\le k<d\).
4. Hence \(a_d \in \{0, \varphi(d)\}\) for each \(d\).
If there is no element of order \(d\), then \(a_d = 0\); if there is one, then the unique subgroup of order \(d\) is cyclic and contains all \(\varphi(d)\) elements of order \(d\), so \(a_d = \varphi(d)\).
5. \(n = \sum_{d \mid n} a_d \le \sum_{d \mid n} \varphi(d) = n\).
By Lagrange’s theorem every element has order dividing \(n\); and \(\sum_{d \mid n} \varphi(d) = n\).
6. Hence \(a_d = \varphi(d)\) for every \(d \mid n\).
4 and 5 force equality term-by-term.
7. In particular, \(a_n = \varphi(n) > 0\), so \(G\) has an element of order \(n\).
6 with \(d = n\).
8. Therefore \(G\) is cyclic.
An element of order \(n\) generates a subgroup of order \(n=|G|\), which is \(G\).
Part 2.
1. Let \(H \le F^\times\) be a finite subgroup of order \(n\).
This introduces notation.
2. For each divisor \(d\) of \(n\), \(H\) has at most one subgroup of order \(d\).
The elements of \(H\) of order dividing \(d\) are roots of the polynomial \(x^d - 1\) in the field \(F\), and a polynomial of degree \(d\) over a field has at most \(d\) roots; hence there are at most \(d\) such elements, so at most one subgroup of order \(d\).
3. Hence \(H\) is cyclic by part 1.
2 and part 1.
4. Q.E.D.
8 (part 1) and 3 (part 2).