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The order of an element of a direct product of groups

  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##A## and ##B## be finite groups, and ##A \times B## be their direct product. Given that ##(a,1)## and ##(1,b)## commute, and that ##(a,1)^n = (a^n,1)## and ##(1,b)^n = (1,b^n)## for all a and b, show that the order of ##(a,b)## is the least common multiple of the orders of a and b.

    2. Relevant equations


    3. The attempt at a solution
    Let ##n## be the order of ##(a,b)##. Then ##(a,b)^n = (1,1)##,
    ##[(a,1)(1,b))]^n = (1,1)##
    ##(a,1)^n (1,b)^n = (1,1)##
    ##(a^n,1)(1,b^n) = (1,1)##
    ##(a^n,b^n) = (1,1)##
    So ##a^n = 1##and ##b^n = 1##. It must be the case that n divides the order of ##a## and the order of ##a##. The smallest number that divides both is by definition the least common multiple.

    Is this proof okay?
     
  2. jcsd
  3. Mar 29, 2017 #2

    fresh_42

    Staff: Mentor

    Almost.
    O.k. until here.
    It must be the case that the order of ##a## and the order of ##b## divide ##n##. The smallest number that is divided by both is by definition the least common multiple.

    Do you know why ##a^{ord(a)}=1=a^n## implies ##n\,\vert \,ord(a)\,##?
     
  4. Mar 29, 2017 #3
    Right, my brain slipped up when I was typing. I know that if ##a^n = 1##, then the order of a must divide ##n##, specifically because the division algorithm
     
  5. Mar 29, 2017 #4
    Right, my brain slipped up when I was typing. I know that if ##a^n = 1##, then the order of a must divide ##n##, specifically because the division algorithm
     
  6. Mar 30, 2017 #5
    Indeed, you cannot have nonzero remainder, for it would contradict the minimality of the order, therefore ##\mbox{ord}(a)\mid n ##, I'm guessing fresh_42 meant it this way. The other way need not hold.
     
  7. Mar 30, 2017 #6

    fresh_42

    Staff: Mentor

    Oops. Slippery ground, indeed.
     
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