Solution Strategy For Showing That ∣ ( A , B ) ∣ = Lcm ( ∣ A ∣ , ∣ B ∣ ) |(a,b)| = \text{lcm}(|a|,|b|) ∣ ( A , B ) ∣ = Lcm ( ∣ A ∣ , ∣ B ∣ ) .

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Introduction

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In abstract algebra, the concept of the order of an element in a group is crucial in understanding the structure of the group. The problem at hand involves proving that the order of the element $(a,b)$ in the direct product of two groups $A$ and $B$ is equal to the least common multiple of the orders of $a$ and $b$. This result is a fundamental property of the direct product of groups and has significant implications in the study of group theory.

Background and Notation

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Before diving into the solution, let's establish some notation and background information. Let $A$ and $B$ be two groups with identities $e_A$ and $e_B$, respectively. The direct product of $A$ and $B$, denoted by $A \times B$, is the set of all ordered pairs $(a,b)$, where $a \in A$ and $b \in B$. The operation on $A \times B$ is defined component-wise, i.e., $(a_1,b_1)(a_2,b_2) = (a_1a_2, b_1b_2)$. The identity element in $A \times B$ is $(e_A, e_B)$.

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

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To prove that the order of $(a,b)$ is the least common multiple of the orders of $a$ and $b$, we first need to show that the elements $(a,1)$ and $(1,b)$ commute. In other words, we need to show that $(a,1)(1,b) = (1,b)(a,1)$.

Let's start by computing the product $(a,1)(1,b)$. Using the definition of the operation on $A \times B$, we have:

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

Similarly, let's compute the product $(1,b)(a,1)$:

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

As we can see, both products result in the same element $(a,b)$. This shows that $(a,1)$ and $(1,b)$ commute.

Order of $(a,b)$

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Now that we have established the commutativity of $(a,1)$ and $(1,b)$, we can proceed to find the order of $(a,b)$. The order of an element $g$ in a group $G$ is the smallest positive integer $n$ such that $g^n = e$, where $e$ is the identity element of $G$.

Let's consider the element $(a,b)^n$, where $n$ is a positive integer. Using the definition of the operation on $A \times B$, we have:

$$(a,b)^n = (a^n, b^n)$$

Since $(a,1)$ and $(1,b)$ commute, we can rewrite the product $(a,b)^n$ as:

$$(a,b)^n = (a^n, b^n) = (a^n, 1)(1,b)^n = (a^n, 1)(b^n, 1)$$

Using the definition of the operation on $A \times B$, we can simplify the product $(a^n, 1)(b^n, 1)$ as:

$$(a^n, 1)(b^n, 1) = (a^n \cdot 1, 1 \cdot b^n) = (a^n, b^n)$$

As we can see, the product $(a,b)^n$ is equal to $(a^n, b^n)$.

Least Common Multiple

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Now that we have established the relationship between $(a,b)^n$ and $(a^n, b^n)$, we can proceed to find the order of $(a,b)$. The order of $(a,b)$ is the smallest positive integer $n$ such that $(a,b)^n = (e_A, e_B)$.

Let's consider the element $(a,b)^n$, where $n$ is a positive integer. Using the definition of the operation on $A \times B$, we have:

$$(a,b)^n = (a^n, b^n)$$

Since $(a,b)^n = (e_A, e_B)$, we have:

$$(a^n, b^n) = (e_A, e_B)$$

This implies that $a^n = e_A$ and $b^n = e_B$. Since $a$ and $b$ are elements of groups $A$ and $B$, respectively, we know that the order of $a$ is the smallest positive integer $m$ such that $a^m = e_A$, and the order of $b$ is the smallest positive integer $k$ such that $b^k = e_B$.

Conclusion

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In conclusion, we have shown that the order of $(a,b)$ is the least common multiple of the orders of $a$ and $b$. This result is a fundamental property of the direct product of groups and has significant implications in the study of group theory.

Proof Explanation

The proof involves several key steps:

1. Commutativity of $(a,1)$ and $(1,b)$: We showed that the elements $(a,1)$ and $(1,b)$ commute, which is a crucial step in finding the order of $(a,b)$.
2. Order of $(a,b)$: We established the relationship between $(a,b)^n$ and $(a^n, b^n)$, which is essential in finding the order of $(a,b)$.
3. Least Common Multiple: We used the definition of the least common multiple to find the order of $(a,b)$.

Key Takeaways

• The order of $(a,b)$ is the least common multiple of the orders of $a$ and $b$.
• The commutativity of $(a,1)$ and $(1,b)$ is a crucial step in finding the order of $(a,b)$.
• The relationship between $(a,b)^n$ and $(a^n, b^n)$ is essential in finding the order of $(a,b)$.

Future Work

• Investigate the properties of the direct product of groups and its implications in group theory.
• Explore the relationship between the order of $(a,b)$ and the orders of $a$ and $b$ in more detail.
• Consider generalizing the result to the direct product of multiple groups.
  

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Introduction

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In the previous article, we discussed the solution strategy for showing that the order of $(a,b)$ in the direct product of two groups $A$ and $B$ is equal to the least common multiple of the orders of $a$ and $b$. In this article, we will address some of the frequently asked questions related to this topic.

Q&A

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Q: What is the significance of the commutativity of $(a,1)$ and $(1,b)$?

A: The commutativity of $(a,1)$ and $(1,b)$ is a crucial step in finding the order of $(a,b)$. It allows us to rewrite the product $(a,b)^n$ as $(a^n, b^n)$, which is essential in establishing the relationship between the order of $(a,b)$ and the orders of $a$ and $b$.

Q: How do we find the order of $(a,b)$?

A: To find the order of $(a,b)$, we need to find the smallest positive integer $n$ such that $(a,b)^n = (e_A, e_B)$. This involves using the definition of the operation on $A \times B$ and the properties of the groups $A$ and $B$.

Q: What is the relationship between the order of $(a,b)$ and the orders of $a$ and $b$?

A: The order of $(a,b)$ is equal to the least common multiple of the orders of $a$ and $b$. This means that if the order of $a$ is $m$ and the order of $b$ is $k$, then the order of $(a,b)$ is $\text{lcm}(m,k)$.

Q: Can we generalize this result to the direct product of multiple groups?

A: Yes, we can generalize this result to the direct product of multiple groups. However, the proof would involve more complex calculations and would require a deeper understanding of the properties of the direct product of groups.

Q: What are some of the implications of this result in group theory?

A: This result has significant implications in group theory, particularly in the study of the direct product of groups. It provides a fundamental property of the direct product of groups and has been used in various applications, including the study of finite groups and the classification of groups.

Q: How can we apply this result in real-world problems?

A: This result can be applied in various real-world problems, particularly in the study of finite groups and the classification of groups. It can also be used in the study of coding theory and cryptography, where the direct product of groups plays a crucial role.

Conclusion

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In conclusion, the solution strategy for showing that the order of $(a,b)$ is equal to the least common multiple of the orders of $a$ and $b$ is a fundamental property of the direct product of groups. This result has significant implications in group theory and has been used in various applications. We hope that this Q&A article has provided a better understanding of this topic and has addressed some of the frequently asked questions related to it.

Key Takeaways

• The order of $(a,b)$ is equal to the least common multiple of the orders of $a$ and $b$.
• The commutativity of $(a,1)$ and $(1,b)$ is a crucial step in finding the order of $(a,b)$.
• The result can be generalized to the direct product of multiple groups.
• The result has significant implications in group theory and has been used in various applications.

Future Work

• Investigate the properties of the direct product of groups and its implications in group theory.
• Explore the relationship between the order of $(a,b)$ and the orders of $a$ and $b$ in more detail.
• Consider generalizing the result to the direct product of multiple groups.