Group Axioms Under Multiplication

A set G forms a group under a binary operation (like multiplication) if it satisfies four conditions:

• Closure: For any two elements a and b in G, the result of the operation a * b must also be in G.
• Associativity: For any elements a, b, and c in G, the equation (a * b) * c = a * (b * c) must hold.
• Identity Element: There must exist an element e in G such that for every element a in G, a * e = e * a = a.
• Inverse Element: For each element a in G, there must exist an element a^-1 in G such that a * a^-1 = a^-1 * a = e (the identity element).

Analyzing the Set $\{1, -1, i, -i\}$

Let the set be $G = \{1, -1, i, -i\}$. We check the group axioms under multiplication:

• Closure: Multiplying any two elements from G results in an element within G. For example, $i \times i = -1$, $(-1) \times i = -i$, and $i \times (-i) = 1$. All products are in G.
• Associativity: Multiplication of complex numbers is associative.
• Identity Element: The multiplicative identity is 1, which is present in G.
• Inverse Element:
  • The inverse of 1 is 1, since $1 \times 1 = 1$.
  • The inverse of -1 is -1, since $(-1) \times (-1) = 1$.
  • The inverse of $i$ is $-i$, since $i \times (-i) = 1$.
  • The inverse of $-i$ is $i$, since $(-i) \times i = 1$.
  All elements have their inverses within G.

Since all four axioms are satisfied, the set $\{1, -1, i, -i\}$ forms a group under multiplication.

Analyzing Other Sets

• Natural Numbers: Does not have multiplicative inverses for most elements (e.g., 2 has no inverse in $\mathbb{N}$).
• Irrational Numbers: Not closed under multiplication (e.g., $\sqrt{2} \times \sqrt{2} = 2$, which is rational) and lacks the identity element 1.
• Rational Numbers: Contains 0, which does not have a multiplicative inverse. The set of *non-zero* rational numbers forms a group, but the set including 0 does not.

Conclusion

The set $\{1, -1, i, -i\}$ is the only set listed that satisfies all the group axioms under multiplication.