Properties of the Direct Product

Oct 13, 2025 by pynkocean.com 40 views

=====================================================

Introduction

In abstract algebra, the concept of the least common multiple (LCM) plays a crucial role in understanding the properties of groups and rings. The problem at hand, as stated in Dummitt and Foote, requires us to prove that the order of the element $(a,b)$ in the direct product $A \times B$ is equal to the least common multiple of the orders of $a$ and $b$ . In this article, we will delve into the solution strategy for showing that $|(a,b)| = \text{lcm}(|a|,|b|)$ .

Understanding the Problem

To begin with, let's break down the problem statement. We are given two elements $(a,1)$ and $(1,b)$ in the direct product $A \times B$ . The problem asks us to prove that these two elements commute, and then deduce that the order of $(a,b)$ is the least common multiple of the orders of $a$ and $b$ . This requires us to understand the properties of the direct product of groups and the concept of the least common multiple.

Properties of the Direct Product

The direct product of two groups $A$ and $B$ , denoted by $A \times B$ , is a group under the operation $(a,b) \cdot (c,d) = (ac,bd)$ . One of the key properties of the direct product is that it preserves the order of the elements. Specifically, if $a \in A$ has order $n$ , then $(a,1) \in A \times B$ also has order $n$ .

Commutativity of $(a,1)$ and $(1,b)$

To prove that $(a,1)$ and $(1,b)$ commute, we need to show that $(a,1) \cdot (1,b) = (1,b) \cdot (a,1)$ . Using the definition of the direct product, we have:

(a,1) \cdot (1,b) = (a \cdot 1, 1 \cdot b) = (a,b)

(1,b) \cdot (a,1) = (1 \cdot a, b \cdot 1) = (a,b)

As we can see, both expressions simplify to $(a,b)$ , which means that $(a,1)$ and $(1,b)$ commute.

Order of $(a,b)$

Now that we have established the commutativity of $(a,1)$ and $(1,b)$ , we can proceed to find the order of $(a,b)$ . Let $n = |a|$ and $m = |b|$ . We know that $(a,1)$ has order $n$ and $(1,b)$ has order $m$ . Since $(a,1)$ and $(1,b)$ commute, we can write:

(a,b)^{nm} = ((a,1) \cdot (1,b))^{nm} = (a,1)^n \cdot (1,b)^m = (1,1)

This shows that $(a,b)^{nm}$ is the identity element in $A \times B$ . Therefore, the order of $(a,b)$ divides $nm$ .

Least Common Multiple

To show that the order of $(a,b)$ is equal to the least common multiple of $n$ and $m$ , we need to prove that $nm$ is the smallest positive integer that divides both $n$ and $m$ . This is a well-known property of the least common multiple.

Conclusion

In conclusion, we have shown that the order of $(a,b)$ in the direct product $A \times B$ is equal to the least common multiple of the orders of $a$ and $b$ . This result is a direct consequence of the commutativity of $(a,1)$ and $(1,b)$ and the properties of the direct product of groups.

Proof Explanation

The proof of this result can be broken down into several steps:

Commutativity of $(a,1)$ and $(1,b)$ : We showed that $(a,1)$ and $(1,b)$ commute by demonstrating that $(a,1) \cdot (1,b) = (1,b) \cdot (a,1)$ .
Order of $(a,b)$ : We found the order of $(a,b)$ by considering the orders of $(a,1)$ and $(1,b)$ and using the fact that they commute.
Least Common Multiple: We proved that the order of $(a,b)$ is equal to the least common multiple of the orders of $a$ and $b$ by showing that $nm$ is the smallest positive integer that divides both $n$ and $m$ .

Final Thoughts

In this article, we have provided a detailed solution strategy for showing that $|(a,b)| = \text{lcm}(|a|,|b|)$ . We have broken down the problem into several steps and provided a clear explanation of each step. We hope that this article has been helpful in understanding the properties of the direct product of groups and the concept of the least common multiple.

References

Dummitt, M., & Foote, R. (2004). Abstract algebra. John Wiley & Sons.

Frequently Asked Questions

In this article, we will address some of the most common questions related to the solution strategy for showing that $|(a,b)| = \text{lcm}(|a|,|b|)$ .

Q: What is the direct product of two groups?

A: The direct product of two groups $A$ and $B$ , denoted by $A \times B$ , is a group under the operation $(a,b) \cdot (c,d) = (ac,bd)$ .

Q: Why is the order of $(a,b)$ equal to the least common multiple of $n$ and $m$ ?

A: The order of $(a,b)$ is equal to the least common multiple of $n$ and $m$ because $(a,b)^{nm}$ is the identity element in $A \times B$ , and $nm$ is the smallest positive integer that divides both $n$ and $m$ .

Q: How do we prove that $(a,1)$ and $(1,b)$ commute?

A: We prove that $(a,1)$ and $(1,b)$ commute by demonstrating that $(a,1) \cdot (1,b) = (1,b) \cdot (a,1)$ .

Q: What is the significance of the least common multiple in this context?

A: The least common multiple is significant in this context because it represents the smallest positive integer that divides both $n$ and $m$ , which is the order of $(a,b)$ .

Q: Can you provide an example to illustrate this concept?

A: Let's consider the groups $A = \mathbb{Z}_4$ and $B = \mathbb{Z}_6$ . The direct product $A \times B$ is a group under the operation $(a,b) \cdot (c,d) = (ac,bd)$ . Let $a = 2 \in A$ and $b = 3 \in B$ . Then, the order of $(a,b)$ is equal to the least common multiple of $2$ and $3$ , which is $6$ .

Q: How does this result relate to the concept of the least common multiple in abstract algebra?

A: This result relates to the concept of the least common multiple in abstract algebra because it shows that the order of an element in a direct product of groups is equal to the least common multiple of the orders of the elements in the individual groups.

Additional Resources

For further reading and exploration, we recommend the following resources:

Dummitt, M., & Foote, R. (2004). Abstract algebra. John Wiley & Sons.
Rotman, J. J. (1999). An introduction to the theory of groups. Springer-Verlag.
Artin, E. (2011). Galois theory. Dover Publications.

Conclusion

In this article, we have addressed some of the most common questions related to the solution strategy for showing that $|(a,b)| = \text{lcm}(|a|,|b|)$ . We hope that this article has been helpful in understanding the properties of the direct product of groups and the concept of the least common multiple.

Final Thoughts

The concept of the least common multiple is a fundamental idea in abstract algebra, and understanding its properties is essential for working with groups and rings. We hope that this article has provided a clear and concise explanation of the solution strategy for showing that $|(a,b)| = \text{lcm}(|a|,|b|)$ .

References

Dummitt, M., & Foote, R. (2004). Abstract algebra. John Wiley & Sons.