1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    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.

    QED


    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 "<=".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof about orders of elements. Ord(a)=ord(bab^-1).
Loading...