Question

Textbook: Algebra A Graduate Course - Isaacs

Prove that the group G has exactly two subgroups iff |G| is prime (and hence finite).

Answer 1

Let G has exactly two subgroups, then they must be the trivial subgroups namely {e} and G itself. Let order of G is finite say n and let any a∈G with a≠e, then 〈a〉=G as G has only two subgroups. Suppose n is not prime, then there exist natural numbers m,q with n=mq But then the element a^m generates a subgroup 〈a^m〉 of order q (because (a^{​​​​​​m})^q=a^{​​​​​​mq}=a^{​​​​​​n}=e). However 1<q<n=o(G). This contradicts that G has only trivial subgroups.
Hence G must be a finite group of prime order .
Conversely let order of G is prime. Then by Lagrange 's theorem order of subgroup must divide the order of group .since order of G is prime say o(G)= p and every prime number p has only two positive divisor 1 and p.
Hence G has only two subgroups of order 1 and p ,i.e. G has only two subgroups namely {e} and G itself. Thus every group of prime order must have exactly two subgroups.