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Math Help - |a\times b| = LCM[|a|, |b|]

  1. #1
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    |a\times b| = LCM[|a|, |b|]

    A and B are groups with binary operations * and ?

    Prove that the order of a\times b is \text{LCM}[|a|,|b|]

    The binary operation of A\times B is define as such:
    (a\times b)(c\times d)=(a*c)\times (b\text{?} d)

    Let |a|=n and |b|=m.

    Do I need to define the gcd as well? Regardless if I do or not, I am not sure where to proceed next.
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  2. #2
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    Re: |a\times b| = LCM[|a|, |b|]

    LCM(n, m) is the smallest integer that is both a multiple of both n and m. By definition the order of a\times b is the least such integer k with (a \times b)^k = (a*\cdots *a) \times (b*\cdots *b) = a^k \times b^k = e \times e. By definition of the order of a and b, we know that a^k = e \Leftrightarrow n | k and b^k = e \Leftrightarrow m | k. Therefore, k is the least such integer such that k = l_an and k = l_bm (these expressions come from definition of n "divides" k, or "n | k"). This means, overall, that k is the least such integer that is a multiple of both n and m.
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