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Math Help - group

  1. #1
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    group

    Hello


    1. G is a group of order p^n, with p prime number and n natural number, which acts on a finite set of cardinal X not divisible by p. Show that there is some element x in X such that gx = x for all g in G.


    2. If p and q are primes with p <q. Prove that every group G of order pq has only one subgroup of order q normal in G. If q is not congruent to 1 modulo p, show that G is abelian and cyclic.


    Thanks, I hope some help
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  2. #2
    GJA
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    Re: group

    Hi, resident911.

    Some tips:

    Problem #1:

    Define

    X^{G}=\{x\in X : gx=x \forall g\in G\}.

    Since G is a p-group, a consequence of the orbit-stabilizer theorem is that

    |X|\equiv |X^{G}| mod p.

    Hopefully this looks familiar from class or a book. If not I can give more details later.

    Now use our assumption on the order of X and see what you can do.

    Problem #2:

    I would suggest looking at the Sylow theorems and seeing where they take you.

    Give it some thought and I'm sure you'll get it!

    Let me know if there's still confusion. Good luck!
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  3. #3
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    Re: group

    one doesn't need the sylow theorems for problem #2:

    by cauchy's theorem, G has a subgroup H of order p, and a subgroup K of order q. since H∩K = {e}, G = HK.

    now [G:K] = p, which means we can let G act on the coset space G/K to get a homomorphism φ:G→Sp.

    consider φ(K). for any element x in φ(K), we have |x| divides q. thus either K is in the kernel of φ, or φ(K) ≅ K.

    but Sp contains no elements of order q, since q does not divide p! (because p < q). so K lies in the kernel.

    so for any k in K, and g in G, k(gK) = gK, so g-1kgK = K, that is: g-1kg is in K.

    thus K is normal in G.

    this means that for any h in H (which is certainly an element of G) hkh-1 is in K.

    in other words, every h in H induces an action on K by conjugation h.k = hkh-1.

    this gives a homomorphism from H to Aut(K), and since K is cyclic of order q, |Aut(K)| = q-1.

    suppose h in H is a generator. then the automorphism k → hkh-1 has order dividing p, and q-1.

    if p does not divide q-1, then the automorphism k → hkh-1 must be the identity automorphism:

    i.e.; hkh-1 = k, for all k in K. in particular, if k is a generator for K, then h and k commute.

    but G = HK, so we have for any g in G, g = hmkn, for some 0 ≤ m < p, 0 ≤ n < q.

    thus gg' = (hmkn)(hm'kn') = (hm+m')(kn+n')

    = (hm'+m)(kn'+n) = (hm'kn')(hmkn) = g'g, G is abelian.

    but if G is abelian, then H is also normal in G, whence G = H x K ≅ Zp x Zq ≅ Zpq, since gcd(p,q) = 1

    (one can also show directly that for a generator h of H, and generator k of K, the element hk has order pq = lcm(p,q)).

    *****************

    to show that a group of order pq where p divides q-1 need not be abelian (let alone cyclic), let p = 2, and q be any odd prime, and consider the dihedral group of order 2q (for q = 3, this is isomorphic to S3).

    *****************

    in all fairness, a proof using the sylow theorems *would* be shorter in showing K is normal in G. but that would not tell one how to show that HK is abelian.
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  4. #4
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    Re: group

    Exercise 2 I stay clear, thanks Deveno.
    In Exercise 1 as I can use the hypothesis on the order of X? To complete the exercise? I have no very clear that
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  5. #5
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    Re: group

    you want to prove that |XG| ≥ 1, right? this is the same as proving |XG| ≠ 0.

    if a = b (mod p) this means:

    a = b + kp.

    suppose p divides b. then p divides b + kp, and thus p divides a.

    now if we already know p does NOT divide a, then p cannot possibly divide b, either. in particular, b cannot be 0.
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  6. #6
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    Re: group

    hi

    understood the exercise, thank you very much.
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