Problem 3

Question

Prove that each of the following sets, with the indicated operation, is an abelian group. $$ x * y=x+y+x y, \text { on the set }\\{x \in \mathbb{R}: x \neq-1\\} $$

Step-by-Step Solution

Verified

Answer

The set with operation forms an abelian group: closed, associative, has identity (0), inverses $\frac{-x}{1+x}$, and is commutative.

1Step 1: Define the Operation and Set

The given operation is $ x * y = x + y + xy $. The set is $ \{ x \in \mathbb{R} : x eq -1 \} $, which means all real numbers except $ -1 $. We need to demonstrate that this set and operation form an abelian group, adhering to group properties: closure, associativity, identity, inverses, and commutativity.

2Step 2: Check Closure

To check closure, we must verify that $ x * y = x + y + xy $ results in an element that is still within the set $ \{ x \in \mathbb{R} : x eq -1 \} $. For any $ x, y \in \mathbb{R} \setminus \{-1\} $, $ x * y $ is also a real number and cannot be $-1$ because neither $ x $ nor $ y $ are $-1$, and the combination $x + y + xy$ won't be $-1$, ensuring closure.

3Step 3: Verify Associativity

To check associativity, compute $(x * y) * z$ and $x * (y * z)$ and show they are equal. Calculating:$(x * y) * z = (x + y + xy) + z + (x + y + xy)z $$(x + y + xy) + z + xz + yz + xyz$$x + y + xy + z + xz + yz + xyz $Meanwhile,$x * (y * z) = x + (y + z + yz) + x(y + z + yz) $$x + y + z + yz + xy + xz + xyz $Both results simplify to $x + y + z + xy + xz + yz + xyz$, proving associativity.

4Step 4: Identify Identity Element

An identity element $ e $ satisfies $ x * e = x $ and $ e * x = x $. Setting $ e = 0 $, calculate:$x * 0 = x + 0 + x(0) = x$$0 * x = 0 + x + 0(x) = x$Thus, $0$ is the identity element for the operation.

5Step 5: Find Inverse Element

An inverse for $ x $ is $ x' $ such that $ x * x' = 0 $ and $ x' * x = 0 $. Set $ x' = \frac{-x}{1+x}$ (as long as $ x eq -1 $, $ 1+x eq 0 $). Check:$x * x' = x + \frac{-x}{1+x} + x\left(\frac{-x}{1+x}\right) $$= x - x + 0 = 0$Therefore, $ x' = \frac{-x}{1+x} $ is the inverse.

6Step 6: Establish Commutativity

For commutativity, verify $ x * y = y * x $.$x * y = x + y + xy$$y * x = y + x + yx$Since addition is commutative, $ x + y + xy = y + x + yx $, confirming commutativity.

Key Concepts

Closure PropertyAssociative PropertyIdentity ElementInverse ElementCommutative Property

Closure Property

The closure property is essential for a set to qualify as a group under a specific operation. This property ensures that the operation performed on any two elements of the set stays within the set itself. In our context, with the operation defined as $x * y = x + y + xy$ and the set being $ \{x \in \mathbb{R} : x eq -1 \}$, we need to confirm that for any $x$ and $y$ in this set, the outcome $x * y$ is also in the set.

First, select any two elements $x$ and $y$ from the set.
Calculate $x * y$ using the given operation.
Verify that the result, $x + y + xy$, is not equal to $-1$, since this remains true for any $x, y eq -1$.

Therefore, the set retains closure under the given operation.

Associative Property

Associativity is another crucial property for groups. It validates that the order in which operations are performed does not change the final result. For the operation $x * y = x + y + xy$, associativity requires that $(x * y) * z = x * (y * z)$ for all elements.

Calculate $(x * y) * z$ which expands to $(x + y + xy) + z + (x + y + xy)z$.
Similarly, calculate $x * (y * z)$ as $x + (y + z + yz) + x(y + z + yz)$.
Simplify both expressions to uncover that they equal $x + y + z + xy + xz + yz + xyz$.

This computation shows that the operation is indeed associative.

Identity Element

An identity element is a special member of a set that leaves other elements unchanged when used in an operation with them. For the operation in question, an identity element $e$ satisfies both $x * e = x$ and $e * x = x$.

Suppose $e = 0$ for this operation.
Verify $x * 0 = x + 0 + x(0) = x$.
Ensure $0 * x = 0 + x + 0(x) = x$.

With both conditions met, it is clear that 0 acts as the identity element for this set and operation.

Inverse Element

The inverse of an element is another element in the set that combines with the first element to yield the identity element of the operation. For each $x$, we find $x'$ such that both $x * x' = 0$ and $x' * x = 0$ hold true, where 0 is the identity element.

Assume the inverse $x' = \frac{-x}{1+x}$, ensuring $x eq -1$.
Calculate $x * x' = x + \frac{-x}{1+x} + x\left(\frac{-x}{1+x}\right)$.
Simplify to show this equals 0, confirming $x'$ is the inverse.

Thus, every element $x$ in the set has an inverse, maintaining the group property of having an inverse for every element.

Commutative Property

Commutativity is a defining characteristic of abelian groups, ensuring the order of elements does not affect the outcome of an operation. For the operation $x * y = x + y + xy$, it is necessary to verify that $x * y = y * x$.

Consider the expression $x * y$ which results in $x + y + xy$.
Also compute $y * x$, naturally equating to $y + x + yx$.
Note that due to the commutative nature of addition, these expressions are equal.

This proves that our operation is commutative, thereby confirming the criteria for an abelian group.

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