GATE CS 1996 Q1.4 — Monoids and Groups

To determine which statement is false, let us analyze each option based on the definition of an abelian group. A set $G$ under a binary operation $*$ is an abelian group if it satisfies closure, associativity, existence of an identity element, existence of inverse elements, and commutativity.

• Option A: The set of rational numbers under addition, $(\mathbb{Q}, +)$, is an abelian group. The identity is $0$, and the inverse of any $x \in \mathbb{Q}$ is $-x \in \mathbb{Q}$. This statement is TRUE.
• Option B: The set of integers under addition, $(\mathbb{Z}, +)$, is an abelian group. The identity is $0$, and the inverse of any $n \in \mathbb{Z}$ is $-n \in \mathbb{Z}$. This statement is TRUE.
• Option C: The set of rational numbers under multiplication, $(\mathbb{Q}, \times)$, is not a group. Specifically, the element $0 \in \mathbb{Q}$ does not have a multiplicative inverse, as there is no rational number $x$ such that $0 \times x = 1$. This statement is FALSE.
• Option D: The set of non-zero real numbers under multiplication, $(\mathbb{R} \setminus \{0\}, \times)$, is an abelian group. The identity is $1$, and the multiplicative inverse of any non-zero real $r$ is $\frac{1}{r}$, which is also a non-zero real. This statement is TRUE.

Since statement C is false, the correct option is C.