Prove that the set of all rational numbers of the form 3n6m, m,n[itex]\in[/itex]Z, is a group under multiplication.
The Attempt at a Solution
For this problem I attempted to show that the given set has 1. an Identity element, 2. each element has an inverse, 3. Closure under multiplication, and 4. Associativity.
1. The identity element is 1
2. The inverse is 3-n6-m
3. Closure: the rationals are closed under multiplication, so closure holds, i.e.
(3m6n)(3k6l) = 3m+k6n+l.
4. Associativity: This is a property of the rationals and holds, i.e.
(3m6n[itex]\ast[/itex]3k6l)[itex]\ast[/itex]3p6q = 3m6n[itex]\ast[/itex](3k6l[itex]\ast[/itex]3p6q)