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Please help me to prove this problem.

  1. Nov 22, 2006 #1

    Aju

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    Prove that if G is a finite group with multiplicative notation, and 'a' is element of G then
    a^ (|G|+1) = a
     
  2. jcsd
  3. Nov 22, 2006 #2

    HallsofIvy

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    |G| is the number of elements in G. Look at a0= e (the group identity), a1, a2, a3, ..., aG. Since there are only |G| elements in G, there must be a duplicate in that list of |G|+1 values. Suppose an= am with n< m. Then am-n= e. Now you finish it: Show that the set of powers of a: a, a2, ..., ak=e, with k= m-n, foms a subgroup of G. It follows that k must be a factor of |G|. What, then is a|G|?
     
  4. Nov 22, 2006 #3

    mathwonk

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    perhaps the op does not know "lagrange's" theorem? (due to gauss)
     
  5. Nov 23, 2006 #4

    HallsofIvy

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    If he doesn't then he surely shouldn't be doing a problem like this!
     
  6. Nov 23, 2006 #5
    I think Lagranges theorem is actually due to Camille Jordan. (Gauss discovered the special case of cyclic groups)
     
  7. Nov 23, 2006 #6

    mathwonk

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    agreed, gauss discovered the special case of subgroups of Z/n, (and I believe of units in Z/n). what i refer to is the proof gauss gave, which is exactly the disjoint cosets method which proves the general case.

    i.e. although in the special case a more special proof is available, gauss gave a general proof which works in the general case.

    i am perhaps more strict than some in my method of attribution, but i decline to attribute a theorem to someone who merely observes that someone else's proof works again in anther case, without providing a new argument.

    in teaching group theory this year i worked out the theory of abelian groups first, and was able to take advantage of commutativity in this proof. I.e. in an abelian group, given to elements, there is another element whose order is the lcm of their orders. this implies the lagrqnge theorem for abelian groups.

    i was not able however to come up with a new argument for lagrange without using the idea, due to gauss, of disjoint cosets. this is a genuine idea, which one appreciate when trying to circumvent it. when i noticed gauss had introduced this diea in his proof, i became convinced the real idea of the argument for lagranges theorem was due to gauss. indeed i cannot see anyhting new at all in the general argument not present in gauss version.

    so more accurately this theorem is perhaps first stated and proved in general by lagrange, but the argument given in books today for this theorem is identical that used by gauss earlier on a special case. (see his disquisitiones for this argument).

    so if one were to share the credit, i would still give most of it to gauss.
     
    Last edited: Nov 23, 2006
  8. Nov 23, 2006 #7

    mathwonk

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    after further reflection, i think it is a contribution to realize that an existing argument has anew application. i have often done this myself, although niot always with knowledge of the prior argument.

    but the later proof should ideally acknowledge the earlier instance of it. since gauss name is almost never mentioend in connection with algranges theorem, i emophasized gauss's priority.

    but the reference to jordan is also interesting to me. i am curiious to read his traite des substitutioins and see for myself whetehr he mentiions gauss. and why is lagrange usually credited?
     
  9. Nov 23, 2006 #8

    mathwonk

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    ah history is indeed afscinating and eye opening. a little research reveals that fermats little theorem is a precursor of lagrange's theorem and fermat died in 1665, over a century before gauss's birth.

    so i should remember to be more careful in making attributions. someone always knows an earlier reference for a seminal idea on a topic. note too the encouraging side that even great innovators are only building a little on what went before.

    so we should read the great mathematicians and try to understand what they have done so well that we see ourselves how to extend it.

    indeed the concept of a gcd in euclid could be said to lie beneath all these concepts for cyclic groups. i.e. a cyclic group is a homomorphic image of Z, and subgroups of subgroups of Z correspond to divisors of generators.tht wasin fact my motive in giving credit to gauss for this proof. i.e. euclidfs method works on abelian groups, and gauss method works on non abelian groups. amd i know of no other methods, so to me gauss made the conceptual leap allowing the theorem to generalize to non abelian groups.

    someone else actually noticed that however.
     
    Last edited: Nov 23, 2006
  10. Nov 24, 2006 #9

    HallsofIvy

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    I wonder if poor Aju is still with us!
     
  11. Nov 24, 2006 #10

    mathwonk

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    Mr Aju, the result we are referring to says that, for an element a of a finite group, the first positive exponenent t such that a^t = 1, where 1 is the identity element, is a divisor of the number of elements of the group.

    Gauss's proof was to observe that one can define a partition of the group into subsets of the same order as follows. one set is the distinct powers a,a^2, a^3,....,a^t = 1. another set is the collection of translates of this first set by any other element g of the group. I.e. ga,ga^2,ga^3,.......,ga^t = g. one repeats this process until no more elements remain. this shows that thw group is a disjoint union of translates of the distinct powers of a. hence the number of these powers is a divisor of the order of the group.

    this is called lagranges theorem, although it is predated by many other similar results, by fermat, gauss, jordan, euclid,.....moses...
     
  12. Nov 24, 2006 #11

    mathwonk

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    if you kniow lagranges theorem your problem follows immediately as follows: a^|G|+1 = a^|G| a = 1a = a. so lagrange is the whole matter.
     
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