Problem 19 - A First Course in Abstract Algebra

Question

Let \(S\) be the set of all real numbers except \(-1\). Define \(*\) on \(S\) by $$ a \cdot b=a+b+a b \text {. } $$ a. Show that * gives a binary operation on \(S\). b. Show that \((S, *)\) is a group. c. Find the solution of the equation \(2+x * 3=7\) in \(S\).

Step-by-Step Solution

Answer

\((S, *)\) is a group, and the solution for \(2+x * 3=7\) is \(x = \frac{2}{3}\).

Step 1: Show * as a Binary Operation on S

To demonstrate * is a binary operation on \( S \), we need to show that for all \( a, b \) in \( S \), the operation \( a \cdot b \) results in another element in \( S \).The formula given for the operation is \( a \cdot b = a + b + ab \). Since \( S \) is the set of all real numbers except \(-1\), we observe that for any real numbers \( a, b \) where neither \( a \) nor \( b \) is \(-1\), the result \( a + b + ab \) is also a real number and cannot equal \(-1\). Therefore, \( a \cdot b \) is in \( S \), so \( * \) is a binary operation on \( S \).

Step 2: Check Associativity in (S, *)

For \( * \) to be associative, it must satisfy \( (a \cdot b) \cdot c = a \cdot (b \cdot c) \) for all \( a, b, c \) in \( S \).Compute \( (a \cdot b) \cdot c \):1. Find \( a \cdot b = a + b + ab \).2. Then find \((a \cdot b) \cdot c = (a+b+ab) + c + (a+b+ab)c = a+b+c+ab+ac+bc+abc\).Compute \( a \cdot (b \cdot c) \):1. Find \( b \cdot c = b + c + bc \).2. Then find \( a \cdot (b \cdot c) = a + (b+c+bc) + a(b+c+bc) = a+b+c+ab+ac+abc\).Since both expressions are equal, * is associative.

Step 3: Identify the Identity Element in (S, *)

An identity element \( e \) in \( S \) satisfies \( a \cdot e = a \) and \( e \cdot a = a \) for all \( a \in S \).Set \( a \cdot e = a + e + ae = a \). Solving yields:\[(a+e+ae = a) \Rightarrow e(1+a) = 0 \Rightarrow e = 0\] since \( a eq -1 \).So, 0 is the identity element in \( S \), since \( a \cdot 0 = a \) and \( 0 \cdot a = a \).

Step 4: Find the Inverse Element in (S, *)

An inverse element \( b \) for \( a \) in \( S \) satisfies \( a \cdot b = e \) where \( e = 0 \), the identity.Set \( a + b + ab = 0 \). Solving for \( b \):\[ b = \frac{-a}{1+a} \]Since \( a eq -1 \), \( b \) remains in \( S \). Therefore, the inverse of \( a \) is \( b = \frac{-a}{1+a} \).

Step 5: Solve the Equation \(2+x * 3=7\) in S

Solve for \( x \) in the equation \( 2 + x + 2x + 3 = 7 \).Simplify the equation:\\[ 2 + x + 2x + 3 = 7 \Rightarrow 3x + 5 = 7 \] \[ \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3} \]Thus, the solution in \( S \) is \( x = \frac{2}{3} \).

Key Concepts

Binary Operation

A binary operation is a fundamental concept in group theory that involves combining two elements to produce another element. For a set with a defined binary operation to be valid, the result must stay within the set. In our case, we define a binary operation \(*\) on the set \(S\), which consists of all real numbers except \(-1\). The operation is given as \(a \cdot b = a + b + ab\).

• For any two elements \(a\) and \(b\) in \(S\), neither being \(-1\), the result \(a + b + ab\) will also be a real number that is not \(-1\).
• This ensures that the operation \(*\) is closed within \(S\), thereby making it a binary operation on the set \(S\).

Associative Property

The associative property is a critical feature for \(S, *\) being a group, which requires that the grouping of operations does not affect the outcome, or mathematically, \((a \cdot b) \cdot c = a \cdot (b \cdot c)\).
To verify this,

• Compute \((a \cdot b) \cdot c\): First, calculate \(a \cdot b = a + b + ab\). Then, perform \((a + b + ab) \cdot c = (a+b+ab) + c + (a+b+ab)c = a+b+c+ab+ac+bc+abc\).
• Next, compute \(a \cdot (b \cdot c)\): Begin with \(b \cdot c = b + c + bc\). Next, calculate \(a \cdot (b + c + bc) = a + (b+c+bc) + a(b+c+bc) = a+b+c+ab+ac+abc\).

Since both expressions are equal, \(*\) satisfies the associative property within the set \(S\).

Identity Element

An identity element in a set with a binary operation is an element that, when combined with any element of the set using the operation, leaves the element unchanged. In our set \(S\), we need an element \(e\) such that \(a \cdot e = a\) and \(e \cdot a = a\) for all \(a\) in \(S\).

• Consider \(a \cdot e = a + e + ae = a\). This simplifies to \(e(1 + a) = 0\). Given that \(a eq -1\), \(e = 0\).
• Therefore, 0 acts as the identity element in \((S, *)\) because \(a \cdot 0 = a\) and \(0 \cdot a = a\) for any element \(a\) in \(S\).

Inverse Element

For a group, every element must have an inverse such that combining the element and its inverse result in the identity element. In \((S, *)\), the identity element is 0, thus for any element \(a\) in \(S\), its inverse \(b\) must satisfy \(a \cdot b = 0\).

• Starting from \(a + b + ab = 0\), solving for \(b\) yields \(b = \frac{-a}{1+a}\).
• Since \(a eq -1\), \(1+a eq 0\), ensuring \(b\) is defined and within \(S\).

Thus, the inverse element of \(a\) in \((S, *)\) is \(b = \frac{-a}{1+a}\), achieving the condition \(a \cdot b = 0\).