Question

Let \(G\) be a group of order \(p q,\) where \(p\) and \(q\) are distinct primes. Suppose \(G\) has a unique subgroup of order \(p\) and a unique subgroup of order \(q\). Show that \(G\) is cyclic,

Short answer

The group \(G\) is cyclic, generated by the element \(x = ab\) with order \(pq\).

Step-by-step solution

Identify Subgroups and Properties

According to Lagrange's theorem, the order of any subgroup of a group must divide the order of that group. Since \(G\) is of order \(pq\), it can have subgroups of order \(p\) and \(q\). We are given that these subgroups are unique.

Find Generators of the Subgroups

A unique subgroup of a particular order is cyclic. Therefore, the subgroup of order \(p\) has an element \(a\) such that \(a^p = e\) where \(e\) is the identity, and the subgroup of order \(q\) has an element \(b\) such that \(b^q = e\).

Consider the Product of Generators

Consider the element \(x = ab\). We need to show that \(x\) generates the group \(G\).

Show \(ab\) Operates Over the Whole Group

If the element \(x = ab\) is of order \(pq\), then \(x^{pq} = (ab)^{pq} = e\). Since the order of \(G\) is \(pq\), this implies \(x\) generates \(G\).

Prove Non-triviality of \(x\)

Consider the possibility that \(x^k = e\) for \(k<pq\). But since the subgroups are unique, \(x^k\) reaching \(e\) before \(pq\) would imply \(x\) stays within one of the subgroups, violating its uniqueness. Hence, no smaller power than \(pq\) reaches \(e\).

Conclusion

Since we have demonstrated that \(x = ab\) has order \(pq\), \(G\) is cyclic, being generated by \(x\).

Key concepts

Lagrange's Theorem

Lagrange's theorem plays a pivotal role in understanding the structure of groups. In group theory, a group is a set equipped with an operation that combines any two elements to form a third element while satisfying four conditions: closure, associativity, identity, and invertibility.

Lagrange's theorem specifically states that, if you have a finite group, the order (or size) of every subgroup divides the order of the entire group.

• If a group has order 12, for example, any subgroup must have an order that is a divisor of 12, such as 2, 3, 4, or 6.

This concept is essential for understanding why, in our exercise, the uniqueness of the subgroups of order \(p\) and \(q\) in a group of order \(pq\) suggests deeper structural features about the group itself. With \(G\) having unique subgroups of orders \(p\) and \(q\), both of these subgroups don’t just divide the order; they uniquely define the group’s configuration, laying the groundwork for proving \(G\) is cyclic.

Unique Subgroups

The exercise indicates that \(G\) has unique subgroups of order \(p\) and \(q\). Uniqueness here means exactly one such subgroup exists for each order. This characteristic greatly influences the overall structure of the group \(G\).

• If a subgroup is unique, it must be normal, meaning any element in the group will leave this subgroup unchanged under a conjugate operation.

This is significant because unique subgroups that are normal hold the property that any element from the group commutes with the elements of these subgroups.
This commutativity becomes critical in exercises like ours because it ensures that elements of \(G\) can be combined in various ways without altering subgroup membership, which is a stepping stone in proving that the group is cyclic.

Group Order

The order of a group is a foundational concept that describes how many elements are in the group. Knowing the group's order, especially when expressed as the product of two distinct primes \(p\) and \(q\), provides clues about its possible subgroups through Lagrange’s theorem.

• The fact that \(G\) has order \(pq\) immediately tells us it can have subgroups whose orders are the divisors of \(pq\): 1, \(p\), \(q\), and \(pq\).

Understanding the group's order, therefore, allows us to predict what subgroups could exist and informs us how the entire group can be formed step by step.
In our example, the uniqueness of the subgroups of order \(p\) and \(q\) suggests no other configurations for subgroups can result, providing a direct path to show \(G\) is cyclic.

Generators of Subgroups

A cyclic group is one where a single element, known as a generator, can be used to produce every element of that group through repeated application of the group operation.

• In a unique subgroup of order \(p\), this generator can be described as an element \(a\) such that \(a^p = e\) where \(e\) is the identity element.
• Similarly, in a subgroup of order \(q\), the generator can be an element \(b\) such that \(b^q = e\).

In exercises involving cyclic groups or groups with unique subgroups, identifying these generators is crucial.
The uniqueness implies that no elements exist that could form different subgroups of these same orders, helping to determine that these generators can be combined in such a way that they cover the entire group.
This idea forms the crux of showing that \(G\) is cyclic when noting that the product of these two elements, \(ab\), forms this much larger structure.