Problem 8 - A Book of Abstract Algebra

Question

Suppose a group \(G\) is generated by two elements \(a\) and \(b\). If \(a b=b a\), prove that \(G\) is abelian.

Step-by-Step Solution

Answer

The group \(G\) is Abelian because the commutative property holds for any elements in \(G\).

Step 1: Understand the definition of an Abelian group

An Abelian group is a group where the commutative property holds for all elements. This means for any elements \(x, y\) in \(G\), the equality \(x \cdot y = y \cdot x\) should hold.

Step 2: Use the given property

We are given that \(a \cdot b = b \cdot a\). This is the commutative property, but only for the specific elements \(a\) and \(b\). We want to show this property holds for any elements in \(G\).

Step 3: Consider arbitrary elements in the group

Since \(G\) is generated by \(a\) and \(b\), any element \(x\) in \(G\) can be written in the form \(x = a^m b^n\) for some integers \(m\) and \(n\). Similarly, let \(y = a^p b^q\) be another element in \(G\).

Step 4: Show commutativity for arbitrary elements

We need to prove \(x \cdot y = y \cdot x\) for our arbitrary elements \(x = a^m b^n\) and \(y = a^p b^q\). Start by expanding both sides: - \(x \cdot y = (a^m b^n)(a^p b^q) = a^m b^n a^p b^q\).- Using the commutative property of \(a\) and \(b\), rewrite as \(a^m a^p b^n b^q = a^{m+p} b^{n+q}\).- Now, consider \(y \cdot x = (a^p b^q)(a^m b^n) = a^p b^q a^m b^n\).- Similarly, this becomes \(a^p a^m b^q b^n = a^{p+m} b^{q+n} = a^{m+p} b^{n+q}\).- Hence, \(x \cdot y = y \cdot x\).

Step 5: Conclusion

Since this commutativity holds for arbitrary elements \(a^m b^n\) and \(a^p b^q\), we conclude that the group \(G\) is Abelian.

Key Concepts

Commutative Property

The commutative property in the context of group theory means that the order in which you combine two elements of the group does not matter. More formally, for a group to be considered Abelian (commutative), it must satisfy the condition

• For any elements \(x\) and \(y\) in the group \(G\), the equation \(x \cdot y = y \cdot x\) holds true.

This is an important trait because it simplifies calculations within the group. You don't have to worry about re-ordering elements when multiplying them; they'll always produce the same result.

In our exercise, we are given the specific condition that \(a \cdot b = b \cdot a\), indicating commutativity for the elements \(a\) and \(b\). This is the crux of proving the entire group's commutativity. If we can show that every element in \(G\) behaves like \(a\) and \(b\), then we prove that the group is Abelian overall.

Group Generation

A group being 'generated' by certain elements means that every element within the group can be expressed as some combination of these generators. In our exercise, the group \(G\) is generated by the elements \(a\) and \(b\), which means:

• Any element \(x\) in \(G\) can be written as \(a^m b^n\)% for integers \(m\) and \(n\).

This property is immensely useful in understanding the structure of the group. By knowing that every element is some form of \(a\) and \(b\), we can systematically approach demonstrating properties for the entire group, like in proving it is Abelian. In practice, focusing on the generators provides a straightforward methodology to test for properties like commutativity, by reducing the problem to these simpler building blocks (\(a\) and \(b\)) of the group.

Element Properties

In group theory, the properties of elements can deeply influence the overall characteristics of the group. When we consider elements like \(a\) and \(b\) that can generate the entire group, understanding their individual properties helps us infer attributes for the group.

• For instance, we know in \(G\), \(a^m \cdot b^n = b^n \cdot a^m\) holds true due to given commutative property between \(a\) and \(b\).
• The exponents \(m\) and \(n\) allow elements of \(G\) to transform flexibly, permitting combinations that still maintain these orderly characteristics of the whole set.

These relationships serve as foundational steps to showing that \(x \cdot y = y \cdot x\) for any arbitrary elements \(x = a^m b^n\) and \(y = a^p b^q\). As we've demonstrated in our solution, once you replace \(x\) and \(y\) with \(a^m b^n\) and \(a^p b^q\), and apply the known commutation rule for \(a\) and \(b\), the property manifests naturally across the group, concluding as Abelian.