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.

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- Sep 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. - Sep 25th 2008, 12:19 PMMoo
Hello,

$\displaystyle (ab)^2)=(ab)(ab)$ and $\displaystyle a^2b^2=aabb$

By associativity of the group operation, we have :

$\displaystyle (ab)(ab)=a(ba)b$ and $\displaystyle a(ab)b$

We have $\displaystyle a(ba)b=a(ab)b$

Since a and b are in a group, there exist $\displaystyle a^{-1}$ and $\displaystyle b^{-1}$ such that $\displaystyle aa^{-1}=a^{-1}a=1$ and $\displaystyle bb^{-1}=b^{-1}b=1$

Multiply, the equation by $\displaystyle a^{-1}$ to the right and by $\displaystyle b^{-1}$ to the left :)

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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.

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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.