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Thread: Group Theory

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    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.
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    Quote Originally Posted by JCIR View Post
    Prove that if (ab)^2=a^2b^2 in a group G, then ab=ba.
    $\displaystyle (ab)^2)=(ab)(ab)$ 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

    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.
    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
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