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Help understanding a group theory proof

  1. Apr 6, 2012 #1
    iam currently studying undergraduate abstract algebra and i have reached to the permutation group topic i understand every thing till now but iam having trouble understanding the proof of

    "IF the identity permutation I of {1,2...n} is represented by m transpositions then m is even"

    I understand that the identity permutation should fix all the number that u put in it so I(1)=1 I(2)=2 etc..


    so the proof is by induction it start with n=2 so I is represented by (12)(12)(12).....

    in which it is easy to see that m should be even since I keeps switching 1with 2 till they are in the same position they were in the beginning and (12) fixes all other numbers

    but then how do u do the induction step??

    I dont even understand how it would work for n=3 :( everything was flowing well but then iam stuck with this...if anyone can help me it would be really appreciated
     
  2. jcsd
  3. Apr 6, 2012 #2

    Fredrik

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    I think it will be easier to do the n=3 case first, and then prove that for all integers p≥2, if the statement holds for n=p, then it holds for n=p+2.

    For the n=3 case, prove that if f is the composition of two swaps, then either f=I or f(k)≠k for all k in {1,2,3}.
     
    Last edited: Apr 6, 2012
  4. Apr 6, 2012 #3
    thnx fredrik for the reply..

    i have though about it again before i read ur reply and it seems i proved without the need for induction iam still having trouble using induction as a proof method sadly :(

    anyways i though for n=2 there are 2 numbers that the Identity permutation should fix or else it wouldn't be Identity ,in this case it is obvious that it must have even numbers of swaps to cancel each swap effect....

    for n=3 there are 3 numbers that I(n) should fix,

    for 1 it must have even number of swaps of the form (1n) to keep 1 in its place or have "another swap" that doesnt contain 1 in the first place so it fix it automatically..

    but the swap that doesnt contain 1 would swap something else 2 or 3 in this case so we must have even number of this "other swap" to cancel its effect on 2 or 3

    and so they must be even

    and so for a composition of transpositions to work as an identity permutation and fix all numbers then they must be even in number..since for every number there will be even swaps to fix that number...Now the problem is HOW DO I WRITE THIS IN A FORMAL WAY
     
  5. Apr 6, 2012 #4
    and i dont understand how proving if f is the composition of two swaps, then either f=I or f(k)≠k would prove that the number of swaps must be even for I(n)

    i understand its either that the two swaps are the same (12)(12),(13)(13),(23)(23) so f=I

    or they are different (12)(23) in this case it doesnt fix 1 or 3 and iam guessing 2 as well
     
  6. Apr 6, 2012 #5

    Fredrik

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    Keep in mind that we're talking about the n=3 case here. A permutation of {1,2,3} can be thought of as a rearrangement of the members of the ordered triple (1,2,3). If you find all the possible rearrangements of (1,2,3) that are the result of two swaps, you will see that none of them is restored to (1,2,3) by a single swap, meaning that there's no composition of 3 swaps that's equal to the identity. That's at least a good start of the proof of the n=3 case.
     
  7. Apr 6, 2012 #6
    ohh i see i might try that and see how would it change with n=4 i think by enough examples i would see how the induction step would work thnx
     
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