Prove that if (ab)^2=a^2b^2 in a group G, then ab=ba.

Prove that the set of all rational numbers of the form 3^m6^n, where m and n are integers is a group under multiplication.

Printable View

- September 25th 2008, 11:37 AMJCIRGroup Theory
Prove that if (ab)^2=a^2b^2 in a group G, then ab=ba.

Prove that the set of all rational numbers of the form 3^m6^n, where m and n are integers is a group under multiplication. - September 25th 2008, 12:19 PMMoo
Hello,

and

By associativity of the group operation, we have :

and

We have

Since a and b are in a group, there exist and such that and

Multiply, the equation by to the right and by to the left :)

Quote:

Prove that the set of all rational numbers of the form 3^m6^n, where m and n are integers is a group under multiplication.

Quote:

1. Closure. For all a, b in G, the result of the operation a • b is also in G.b[›]

2. Associativity. For all a, b and c in G, the equation (a • b) • c = a • (b • c) holds.

3. Identity element. There exists an element e in G, such that for all elements a in G, the equation e • a = a • e = a holds.

4. Inverse element. For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element.