Question

Which of the following set forms a group under multiplication:

• A. $\{1, -1, i, -i\}$
• B. The set of natural numbers.
• C. The set of irrational numbers.
• D. The set of rational numbers.

Hint:

Remember that the set of $n$-th roots of unity $U_n = \{e^{2k\pi i / n} \mid k = 0, 1, \dots, n-1\}$ always forms a finite abelian group under complex multiplication. For $n=4$, $U_4 = \{1, i, -1, -i\}$.

Answer 1

The Correct Option is A

Step 1: Concept
In abstract algebra, an algebraic structure $(G, \cdot)$ consisting of a set $G$ together with a binary operation $\cdot$ is called a group if it satisfies four fundamental axioms:
1. Closure: For all $a, b \in G$, the product $a \cdot b \in G$.
2. Associativity: For all $a, b, c \in G$, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
3. Identity Element: There exists an element $e \in G$ such that $a \cdot e = e \cdot a = a$ for all $a \in G$. Under multiplication, $e = 1$.
4. Inverse Element: For every element $a \in G$, there exists an element $a^{-1} \in G$ such that $a \cdot a^{-1} = a^{-1} \cdot a = e$.

Step 2: Key Formulas and Approach
To check if a finite set $G = \{1, -1, i, -i\}$ under complex multiplication forms a group, we verify closure, identity, and inverses.
For infinite sets like $\mathbb{N}$, $\mathbb{I}$ (irrationals), and $\mathbb{Q}$, we check if every element possesses a multiplicative inverse in the set.

Step 3: Step-by-step Explanation

• Testing Option (A): Consider the set $G = \{1, -1, i, -i\}$.
- Closure:
$1 \cdot (-1) = -1 \in G$
$i \cdot i = -1 \in G$
$i \cdot (-i) = 1 \in G$
$(-1) \cdot (-i) = i \in G$
The multiplication table reveals that all entries lie inside $G$.
- Associativity: Multiplication of complex numbers is associative.
- Identity: The element $1 \in G$ serves as the multiplicative identity since $1 \cdot a = a$ for all $a \in G$.
- Inverses:
The inverse of $1$ is $1 \in G$.
The inverse of $-1$ is $-1 \in G$.
The inverse of $i$ is $-i \in G$ because $i \cdot (-i) = 1$.
The inverse of $-i$ is $i \in G$ because $(-i) \cdot i = 1$.
Since all four properties hold, $G = \{1, -1, i, -i\}$ is a multiplicative group (specifically, the cyclic group of 4th roots of unity, $C_4$).

• Testing Option (B): The set of natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$.
The element $2 \in \mathbb{N}$ has no multiplicative inverse in $\mathbb{N}$, because $\frac{1}{2} \notin \mathbb{N}$. Hence, $\mathbb{N}$ is not a group.

• Testing Option (C): The set of irrational numbers.
Closure fails because $\sqrt{2} \cdot \sqrt{2} = 2$, which is rational. Hence, the set of irrationals is not closed under multiplication.

• Testing Option (D): The set of rational numbers $\mathbb{Q}$.
The set $\mathbb{Q}$ includes $0$. The element $0$ has no multiplicative inverse because division by zero is undefined. Therefore, $\mathbb{Q}$ under multiplication is not a group (though $\mathbb{Q} \setminus \{0\}$ is a group).

Step 4: Final Answer
The set $\{1, -1, i, -i\}$ satisfies all four group axioms under multiplication. Thus, Option (A) is the correct answer.