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Regarding a Group Theory Proof on the Order of Elements in a Group

  1. Aug 1, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that for a finite group A, the order of any element in A divides the order of A.

    2. Relevant equations

    The order of an element a of a group A is the smallest positive interger n such that an = 1.

    3. The attempt at a solution

    Well, I know that the order of a finite group A is the number of elements in A. I realize that the statement can be written as "if group A is finite, then the order of an element in A divides the order of A, which is set up nicely for a direct proof. Divisibility would entail that there exist some integer q such that for all elements of A, ord(a)q = ord(A). However, at this point, I don't have an idea as to how to carry out the proof. Any help would be greatly appreciated!
     
  2. jcsd
  3. Aug 1, 2012 #2
    Two quick ideas for you to think about ....

    1. Perhaps you may want to look at the cyclic subgroup generated by an arbitrary element (not the identity) of the group A.

    or...

    2. Maybe divide the order of the finite group A by the order of an element in A and see what happens. [Using the division algorithm]

    :)

    Hope this helps!
     
    Last edited: Aug 1, 2012
  4. Aug 1, 2012 #3
    I just reread your post....

    To use the second approach (which is the one you want), suppose the order of the finite group A is m. Then suppose the order of some element in A, call it a, is n. Then divide m by n using the division algorithm. What do you know about an and am? Then, think about the condition we have on our remainder in the division algorithm. That is, 0≤r<n.
     
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