Define group with an example. Let S = {-1, 1}, define * on S by a*b = a + b + ab. Then: (a) Show that * gives a binary operation on S. (b) Show that (S, *) is a group. (c) Find... Define group with an example. Let S = {-1, 1}, define * on S by a*b = a + b + ab. Then: (a) Show that * gives a binary operation on S. (b) Show that (S, *) is a group. (c) Find the solution of equation 2*x*3 = 7 in S. Read more

Steps to Solve

1. Define a Group with an example

A group $(G, )$ is a set $G$ together with a binary operation $$ that satisfies the following four axioms:

• Closure: For all $a, b \in G$, $a * b \in G$.
• Associativity: For all $a, b, c \in G$, $(a * b) * c = a * (b * c)$.
• Identity: There exists an element $e \in G$ such that for all $a \in G$, $a * e = e * a = a$.
• Inverse: For each $a \in G$, there exists an element $a^{-1} \in G$ such that $a * a^{-1} = a^{-1} * a = e$.

Example: The set of integers $\mathbb{Z}$ with the operation of addition $+$ forms a group $(\mathbb{Z}, +)$. The identity element is 0, and the inverse of any integer $a$ is $-a$.

2. Verification of binary operation on S

We are given $S = \mathbb{R} \setminus {-1}$, and $a * b = a + b + ab$. For the binary operation to hold, $ab$ must be in $S$ for all $a, b \in S$. We have to show that for any $a, b \in S$, $ab \neq -1$.

Assume $ab = -1$. Then $a + b + ab = -1$. Adding 1 to both sides we get $a + b + ab + 1 = 0$. Factoring this gives $(a+1)(b+1) = 0$. So either $a = -1$ or $b = -1$. But $a, b \in S = \mathbb{R} \setminus {-1 }$, so $a \neq -1$ and $b \neq -1$. This is a contradiction; therefore $ab \neq -1$. Thus $*$ gives a binary operation on $S$.

3. Prove Associativity

We need to show that $(ab)c = a(bc)$ for all $a, b, c \in S$. $(ab)c = (a+b+ab)c = (a+b+ab) + c + (a+b+ab)c = a + b + ab + c + ac + bc + abc $ $a(bc) = a(b+c+bc) = a + (b+c+bc) + a(b+c+bc) = a + b + c + bc + ab + ac + abc $ Since both expressions are equal, the associative property holds.

4. Find the Identity Element

We need to find an element $e \in S$ such that $a * e = e * a = a$ for all $a \in S$. Let's assume $a * e = a$. Then $a + e + ae = a$. Subtracting $a$ from both sides gives $e + ae = 0$. Factoring out $e$ gives $e(1+a) = 0$. Since $a \neq -1$, $1+a \neq 0$, so we must have $e = 0$. Since $0 \in S$, $e = 0$ is the identity element.

5. Find the Inverse Element

For each $a \in S$, we need to find an element $a^{-1} \in S$ such that $a * a^{-1} = a^{-1} * a = e = 0$. Let's assume $a * a^{-1} = 0$. Then $a + a^{-1} + aa^{-1} = 0$. Solving for $a^{-1}$ gives $a^{-1}(1+a) = -a$. So $a^{-1} = \frac{-a}{1+a}$. Since $a \neq -1$, the denominator is non-zero. Now we need to check if $a^{-1} \in S$, i.e., $a^{-1} \neq -1$. If $a^{-1} = -1$, then $\frac{-a}{1+a} = -1$. This implies $-a = -(1+a)$, so $-a = -1 - a$, which means $0 = -1$, a contradiction. Therefore, $a^{-1} \neq -1$, and so $a^{-1} \in S$.

Since all four group axioms hold, $(S, *)$ is a group.

6. Solve the equation $2 * x * 3 = 7$

We have $2 * x * 3 = 7$. First, let's simplify $2 * x$: $2 * x = 2 + x + 2x = 2 + 3x$. Now we have $(2*x)*3 = (2+3x)*3 = 7$. So $(2+3x) + 3 + (2+3x)(3) = 7$. $2 + 3x + 3 + 6 + 9x = 7$. $11 + 12x = 7$. $12x = 7 - 11 = -4$. $x = \frac{-4}{12} = -\frac{1}{3}$. Since $-\frac{1}{3} \in S$, the solution is $x = -\frac{1}{3}$.