Let G 12m 12n m n Z Show that G is a group under ordinary

Let G :={ (1+2m)/ (1+2n) : m; n Z Show that G is a group under ordinary multiplication of real numbers

Note: the subject is abstract algebra 2nd edition charles c. printer I got answered from chegg but incorrect I submit wrong answer in the comment please answered depends on subtract algebra math university

Solution

A group G is a non-empty set G together with a binary operation called multiplication (or a product denoted by *) that obeys the following axioms:

(1) a, b G implies a*b G (closure);

(2) a, b, c G implies (a*b)*c = a*(b*c) (associativity);

(3) There exists e G such that a*e = e*a = a for all a G (identity);

(4) For each a G, there exists a^-1 G such that a*a^-1 = a^-1 * a = e

Let G = (1+2m)/ (1+2n) where m; n Z

We will examine whether G satisfies the four prerequisites of a group:

(1 + 2m)/( 1 + 2n)}* [ { ( 1 + 2p)/(1 + 2q)} * {(1 +2d)/(1 + 2e)}] = a* (b * c) as ordinary multiplication is associative.

3. Since 0 Z, (1 + 2*0)/(1 + 2*0) = 1 G. Now if a = (1 + 2m)/ ( 1 + 2n) G, then a * 1 = 1*a = (1 + 2m)/ ( 1 + 2n) = 1. Thus 1, the multiplicative identity belongs to G.

4. Let m, n Z. Then a = (1 + 2m)/(1 +2n) G. Then a^-1 = ( 1 +2n)/(1+2m) also G. Then, we have a*a^-1 = [(1 + 2m)/(1 +2n)] * [( 1 +2n)/(1+2m) ] = 1 and also,   a^-1 * a = [( 1 +2n)/(1+2m)] * [(1 + 2m)/(1 +2n)] = 1 so that a*a^-1 = a^-1 * a = 1.

This proves that G is a group.