The set {1, 2, 3, 5, 7, 8, 9} under multiplication modulo 10 is not a group.…

Concept

A set with an operation is a group only if it satisfies all the group axioms: closure, associativity, an identity element, and an inverse for every element. Failing even one axiom is enough to disqualify it. For multiplication modulo n, a key fact is that an element a has a multiplicative inverse modulo n exactly when gcd(a, n) = 1 (a is coprime to n).

Application

Here n = 10, so we test the given statements against these axioms.

1. Closure: take 2 and 5 from the set. 2 × 5 = 10 ≡ 0 (mod 10), and 0 is not in the set. So the set is not closed under multiplication modulo 10.

2. Inverse criterion: an element has an inverse modulo 10 only when it is coprime to 10. Compute gcd with 10: gcd(2, 10) = 2 and gcd(8, 10) = 2, so neither 2 nor 8 is coprime to 10, and hence neither has a multiplicative inverse modulo 10.

3. The element 3: gcd(3, 10) = 1, so 3 is coprime to 10 and must have an inverse. Indeed 3 × 7 = 21 ≡ 1 (mod 10), so 7 is the inverse of 3 modulo 10.

Cross-check

The question asks for the one reason that is false. The failures of closure, and the lack of inverses for 2 and 8, are all genuine reasons the structure is not a group. The only statement that does not hold is the claim that 3 has no inverse, because 3 × 7 ≡ 1 (mod 10) exhibits its inverse. Hence that is the false reason.