# Group Theory

• September 25th 2008, 11:37 AM
JCIR
Group 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 PM
Moo
Hello,
Quote:

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

$(ab)^2)=(ab)(ab)$ and $a^2b^2=aabb$
By associativity of the group operation, we have :
$(ab)(ab)=a(ba)b$ and $a(ab)b$

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

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

Multiply, the equation by $a^{-1}$ to the right and by $b^{-1}$ 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.
Check these, it's not that difficult ;)