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Math Help - Sylow's theorem

  1. #1
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    Sylow's theorem

    Hi,

    Please can someone help me with the following problems:

    a and b are distinct primes such that a<b, and H is a finite group such that
    l H l = ab

    1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).

    2) Describe explicitly all homomorphisms h: C5 --> Aut(C7), and from this, describe all groups of order 35. How many such subgroups are there?

    3) Same as 2) but this time h: C3 --> Aut(C13), and from this, describe all groups of order 39. Again, how many such subgroups are there?
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  2. #2
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    Quote Originally Posted by jedoob View Post

    1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).
    This statement is not true!

    The theorem says if |G|=pq with q>p, and q is not congruent to 1 modulo p then |G| is a cyclic group.
    However, when it is congruent to 1 modulo p we cannot conlude that.

    However we can show that |G| is not a simple group.
    -Proof

    Let |G|=pq and p>q.
    Now by Sylow's First Theorem there exists a Sylow p-subgroup. Let S(p) denote the number of Sylow p-subgroups. Then we know that by Sylow's Third Theorem:
    S(p) = 1 (mod p) and S(p)|(pq).
    Thus, the statement S(p)|(pq) states that,
    S(p)=1,p,q,pq
    By checking each one we see that,
    p,q,pq do not work in the congruence (because p > q).
    Thus, S(p)=1.
    Now, any inner automorphism under any element maps a Sylow p-subgroup into a Sylow p-subgroup. Since the number of Sylow p-subgroups is fixed at one, it means that the Sylow p-subgroup remains invariant under conjugation, i.e. it is a normal subgroup of G, and hence G is not simple.
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  3. #3
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    Quote Originally Posted by jedoob View Post
    1) How can I use Sylows theorem to show that H has a normal subgroup K with K being isomorphic to Cp (C is the cyclic group).
    sorry i meant K being isomorphic to Cb (not Cp as quoted)
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  4. #4
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    Quote Originally Posted by jedoob View Post
    sorry i meant K being isomorphic to Cb (not Cp as quoted)
    Simple (get the pun) we already shown that there exists a Sylow p-subgroup (p is prime), and there is only one group up to isomorphism of order p, that is, the cyclic group. Proof complete.
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  5. #5
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    Thanks.

    Could you possibly help me with parts 2 and 3 as well?
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