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Homework Help: Proof about orders of elements. Ord(a)=ord(bab^-1).

  1. Jun 16, 2012 #1
    Prove ord(a) = ord (bab-1).

    Attempt at the proof.


    Let G be a group and let a,b ε G.

    Suppose ord(a)=n. That is, an=e. We wish to that the ord(bab-1)=n. If we take an=e and multiply on the left by b and on the right by b-1 then, we have,

    (1) an=e
    (2) banb-1=beb-1
    (3) banb-1=e.
    Hence, we have the fact that,
    (4) an=e=banb-1.
    Now, if we expand (bab-1)n, we get,
    (5) bab-1bab-1bab-1..bab-1=banb-1.
    But, banb-1 = e by our algebra in (1) - (4). Thus, (bab-1)n=banb-1 = e and ord(bab-1)=n.

    Let ord(bab-1)=n. That is, (bab-1)n=e. But, as it was shown in (5), (bab-1)n=banb-1 = e. Using the fact that banb-1 = e, we multiply on the left by b-1 and on the right by b and obtain that an=e. Hence, ord(a)=n.


    How does it look? Thanks!!
  2. jcsd
  3. Jun 16, 2012 #2
    First of all, you can reduce everything you have done in "=>" to one line basically:

    (bab-1)n=banb-1 = beb-1 (since ord a = n) = e.

    But be aware that this has not yet shown ord(bab-1) = n. ord g is defined to be the smallest non-negative integer such that ...

    Same goes for "<=".
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