Introduction:

The study of group theory in mathematics centers on the analysis of sets equipped with a binary operation. A fundamental question in this domain is to characterize groups based on the number of their subgroups. This article delves into groups that possess exactly three subgroups and establishes the relationship between such groups and cyclic groups of order ( p^2 ), where ( p ) is a prime number.

Understanding Subgroups of a Cyclic Group of Order ( p^2 ):

Cyclic Group of Order ( p^2 ):

Let ( G ) be a cyclic group of order ( p^2 ), where ( p ) is a prime. The structure of ( G ) is characterized by its subgroups, which correspond to the divisors of the order ( p^2 ).

Divisors and Subgroups: The divisors of ( p^2 ) are 1, ( p ), and ( p^2 ). Therefore, the subgroups of ( G ) can be explicitly described as follows:

Trivial subgroup: ( {e} ) A subgroup of order ( p ), which is unique due to ( p ) being prime. The group ( G ) itself, having order ( p^2 ).

Thus, ( G ) has exactly three subgroups, confirming the statement that a group has exactly three subgroups if and only if it is cyclic of order ( p^2 ).

Returns to the Question:

Groups with Exactly Three Subgroups: If a group ( G ) has exactly three subgroups, they can be denoted as ( H_1 ), ( H_2 ), and ( G ), where ( H_1 ) is the trivial subgroup and ( H_2 ) is a non-trivial subgroup. By Lagrange's Theorem, the order of any subgroup must divide the order of the group. Let's denote the order of ( G ) as ( n ), and the subgroups must correspond to the divisors of ( n ).

Prime Power and Order Constraints: If ( G ) has exactly three subgroups, then ( n ) can have only three divisors: 1, ( H_2 ), and ( n ). This implies that ( n ) must be a prime power, ( p^k ) for some integer ( k ).

Special Case of ( k 2 ): The smallest non-trivial prime power form is ( p^2 ). This form ensures that ( G ) has exactly three subgroups, as any other prime power or different integer values would introduce additional subgroups.

Proof:

Forward Direction: If ( |G| p^2 ) and ( G ) is cyclic, it has exactly three unique subgroups: the trivial subgroup, a subgroup of order ( p ), and the group ( G ) itself. This is a straightforward consequence of the structure of cyclic groups.

Converse Proof: Suppose ( G ) is a group having exactly three subgroups. Let ( H ) be a subgroup of ( G ) other than the trivial subgroup ( {e} ) and the group ( G ). Since ( H ) is a self-conjugate subgroup, the quotient group ( G / H ) cannot have any non-trivial proper subgroups. Therefore, the order of ( G / H ) must be 1 or ( p ), implying that ( H ) has index ( p ) in ( G ).

By Cauchy's Theorem, if ( G ) has a subgroup of order ( q ) for a prime ( q ) other than ( p ), then ( G ) would have at least three subgroups, which contradicts the assumption. Thus, ( p ) is the only prime dividing the order of ( G ).

Let ( |G| p^n ). Since ( H ) is a maximal subgroup of ( G ), ( |H| p^{n-1} ), and since ( H ) cannot have any non-trivial proper subgroups, we have ( n-1 1 ), leading to ( n 2 ). Therefore, ( |G| p^2 ).

Since ( G ) is finite, and ( p ) is the only prime dividing the order of ( G ), ( G ) must be abelian. Given that ( H ) is a unique subgroup of order ( p ), ( G ) must be cyclic, otherwise, each element of order ( p ) would contradict the uniqueness of ( H ).

Conclusion:

A group having exactly three subgroups is equivalent to being a cyclic group of order ( p^2 ) for some prime ( p ). This characterization is both necessary and sufficient.

Further Reading and Resources:

[1] Abstract Algebra: Theory and Applications by Thomas W. Judson

[2] Subgroups of Cyclic Groups