Question

Let \(G\) be a group with \(|G|>1\) such that \(G\) has no nontrivial proper subgroups. Show that \(G\) is a finite cyclic group of prime order.

Short answer

\( G \) is a finite cyclic group of prime order.

Step-by-step solution

Understanding the Given Properties

We are given a group \( G \) such that \(|G| > 1\) and \( G \) has no nontrivial proper subgroups. This means \( G \) includes only the subgroup containing just the identity and \( G \) itself.

Connecting to Group Order

A group \( G \) having no nontrivial proper subgroups implies that any subgroup generated by an element must either be the trivial group or the whole group. If \( g \) is an element of \( G \), the subgroup created \( \langle g \rangle \) equals \( G \), as no nontrivial proper subgroups exist.

Analyzing Cyclic Nature

Since \( \langle g \rangle = G \) for any non-identity element \( g \), the group \( G \) is cyclic, being generated by any non-identity element itself.

Determining Prime Order

For \( G \) to have only \( \{e\} \) and \( G \) itself as subgroups, the order of \( G \), denoted as \(|G|\), must have no divisors other than 1 and itself. This condition is only satisfied when \(|G|\) is a prime number.

Key concepts

Cyclic Groups

In the world of group theory, cyclic groups hold a special place as some of the simplest yet extremely important groups. A cyclic group is one that can be generated by a single element. This means you can generate every element of the group by combining this particular element with itself repeatedly, using the group operation. Let's consider a group element, say, \( g \). If you can write every other element of the group as \( g^n \) (where \( n \) is an integer), then the group is cyclic.

To visualize this, imagine the group is like a necklace where each bead represents an element, and you only need one bead to create the whole necklace by replication. This simplicity means cyclic groups are easier to handle and understand. The group of integers under addition \( \mathbb{Z} \) is a classic example of an infinite cyclic group where the element 1 generates the entire group, since any integer can be expressed as a sum of 1s and \(-1\)s.

When it comes to finite groups, elements and their orders directly tell you how many times you need to "reapply" an element to get back to the starting point or identity element of the group. If your finite group is cyclic, finding the generator can significantly simplify solving group-related problems, giving you an edge in analysis and understanding.

Subgroups

In mathematical terms, a subgroup is a subset of a group that is itself a group under the same operation. For a subset to form a subgroup, it must satisfy four primary conditions:

• It includes the identity element of the parent group \( G \).
• It is closed under the group operation (meaning if you take two elements in the subgroup and apply the group operation, the result is still in the subgroup).
• It must include the inverse of each of its elements.
• It satisfies associativity, but this is inherently satisfied since the parent group is associative.

If a group has no nontrivial proper subgroups, much like the group \( G \) in our example, it means the only subsets that satisfy these conditions are the trivial group \( \{e\} \) and the group \( G \) itself.

The absence of proper nontrivial subgroups simplifies the group's structure considerably, as the group becomes easier to examine, often reducing down to key investigations on its order and how its elements interact. Observing these properties directly leads us into understanding the group's nature, like whether it's cyclic or has a prime order.

Prime Order

When we describe a group as having prime order, we simply mean the total number of elements in the group, denoted by \(|G|\), is a prime number. A prime number is greater than 1 and has no divisors other than 1 and itself, like 2, 3, 5, 7, and so on.

In terms of subgroups, if a group's order is prime, the Lagrange's theorem in group theory simplifies immensely. This theorem states that the order of every subgroup of \( G \) must be a divisor of the order of \( G \). Consequently, with prime order, there can only be two divisors—1 and the prime number itself. This inevitably means that the only possible subgroups are the trivial subgroup \( \{e\} \) and the group \( G \) itself.

This fact is crucial because it implies that any group with prime order is simple in structure, automatically putting it into the category of cyclic groups. Hence, every element (except the identity) can generate the whole group, making studies or problems involving such groups straightforward to tackle.