Problem 4 - A Book of Abstract Algebra

Question

The symbol \(\mathbb{R} \times \mathbb{R}\) represents the set of all ordered pairs \((x, y)\) of real numbers. \(\mathbb{R} \times \mathbb{R}\) may therefore be identified with the set of all the points in the plane. Which of the following subsets of \(\mathbb{R} \times \mathbb{R}\), with the indicated operation, is a group? Which is an abelian group? \((a, b) *(c, d)=(a c-b d, a d+b c)\), on the set \(\mathbb{R} \times \mathbb{R}\) with the origin deleted.

Step-by-Step Solution

Answer

The subset \\(\mathbb{R} \times \mathbb{R}\backslash \{(0, 0)\}\) is a group and an abelian group.

Step 1: Understand the Group

A group is a set combined with an operation that satisfies four properties: closure, associativity, identity element, and existence of inverses.

Step 2: Define Subset and Operation

The subset in question is \(\mathbb{R} \times \mathbb{R}\) with the origin deleted, \((0,0)\), and the operation is \((a, b) *(c, d) = (a c - b d, a d + b c)\).

Step 3: Check Closure

Closure means that if \( (a, b) \) and \( (c, d) \) are in the set, then so is \( (a c - b d, a d + b c) \). Since the operation results in another pair of real numbers and the origin is avoided, the set is closed.

Step 4: Check Associativity

Associativity requires \[ ((a, b)*(c, d))* (e, f) = (a, b)* ((c, d)*(e, f)). \] Calculating both sides, they simplify in the same way, ensuring that the operation is associative.

Step 5: Find the Identity Element

An identity element is such that \((a, b) * e = (a, b)\). Here, \((1, 0)\) acts as the identity because \((a, b) * (1, 0) = (a \cdot 1 - b\cdot0, a\cdot0 + b\cdot1) = (a, b)\).

Step 6: Check for Inverses

Every element \( (a, b) \) must have an inverse \( (x, y) \) such that \( (a, b) * (x, y) = (1, 0) \). Solving gives \( x = \frac{a}{a^2 + b^2}, y = -\frac{b}{a^2 + b^2} \). Since \((x, y)\) exists for all \( (a, b) \, (0, 0) \) is not present, inverses exist.

Step 7: Determine if the Group is Abelian

A group is abelian if the operation is commutative, i.e., \((a, b) * (c, d) = (c, d) * (a, b)\). Checking both sides, we see \(a c - b d = c a - d b \) and \(a d + b c = c b + d a\), which shows the operation is commutative (abelian).

Key Concepts

Associative Property

The Associative Property is all about how you can group elements when performing operations without changing the result. If you imagine three numbers—let's say, apples, bananas, and cherries—the order in which you group these fruits when adding them doesn’t matter. If \((a, b)*(c, d) = (a c-b d, a d+b c)\), then the property requires that doing the operation \(((a, b)*(c, d))*(e, f)\) yields the same as \((a, b)*((c, d)*(e, f))\). This holds true for our given operation after you simplify both sides.
When verified in the context of our operation on the set of ordered pairs, numerically simplifying both sides gives the same result, confirming associativity.

• This means that no matter how you choose to group the elements, the final outcome remains unchanged.

Identity Element

An identity element in the context of a group operation is a unique element which, when combined with any element of the set, returns that element unchanged. Think of it like a secret ingredient that doesn’t alter the flavor of your dish.
For example, the number 0 is the identity for addition because adding zero doesn’t change the number. Similarly, for multiplication, the identity is 1, because multiplying by 1 leaves the number the same.
In our operation on ordered pairs, \((1, 0)\) acts as the identity element. When you perform the operation \((a, b) * (1, 0)\), it results in \((a, b)\).

• This shows that \((1, 0)\) maintains the value of anything it interacts with under this operation, fulfilling the identity element’s role perfectly.

Abelian Group

A group is called abelian if it satisfies the property of commutativity. This means that the order in which you apply the operation doesn’t alter the result. Imagine swapping seats on a really comfortable sofa—no matter who sits where, the peace and comfort remain.
If for any elements \((a, b)\) and \((c, d)\), \((a, b) * (c, d) = (c, d) * (a, b)\), this operation is commutative.
We calculated \(a c - b d\) and found it to be equal to \(c a - d b\), and further with \(a d + b c = c b + d a\), ensuring our operation behaves well under this swap.

• Thus, every member of the set enjoys the arrangement and agrees on the result of their operation with others, confirming our set is an abelian group.

Inverse Elements

To have an inverse in a group means there’s a buddy back home who can negate or undo your position in the group. This buddy is called your inverse. Imagine you spill salt in a dish, and the right amount of sugar perfectly balances it out.
For every element \((a, b)\), we must find another element \((x, y)\) such that the operation brings you back to normal, i.e., the identity element \((1, 0)\). With the operation \((a, b) * (x, y) = (1, 0)\), solving gives us \(x = \frac{a}{a^2 + b^2}\) and \(y = -\frac{b}{a^2 + b^2}\).
Because \((0, 0)\) is not part of our set (avoiding division by zero), every element can find its matching pair to reverse its effect.

• This ensures each member has a counterpart to restore the balance, securing the inverse property for all elements in the set.