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Math Help - Abstract Algebra - Normal groups

  1. #1
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    Abstract Algebra - Normal groups

    I'll be delighted to receive some guidance in the following questions:

    1. Let G1,G2 be simple groups. Prove that every normal non-trivial subgroup of G= G1 x G2 is isomorphic to G1 or to G2...

    2. Prove that every group of order p^2 * q where p,q are primes is solvable...


    I've no idea about the second question. But the first one seems very easy, but I can't figure out how to solve it.


    Help is needed

    Thanks in advance
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by WannaBe View Post
    I'll be delighted to receive some guidance in the following questions:

    1. Let G1,G2 be simple groups. Prove that every normal non-trivial subgroup of G= G1 x G2 is isomorphic to G1 or to G2...

    2. Prove that every group of order p^2 * q where p,q are primes is solvable...


    I've no idea about the second question. But the first one seems very easy, but I can't figure out how to solve it.


    Help is needed

    Thanks in advance
    For the first one, consider that if N\lhd G_1\times G_2 then the set of first coordinates are invariant under conjugation, and so are the second...soo
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  3. #3
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    Quote Originally Posted by WannaBe View Post
    I'll be delighted to receive some guidance in the following questions:

    1. Let G1,G2 be simple groups. Prove that every normal non-trivial subgroup of G= G1 x G2 is isomorphic to G1 or to G2...

    2. Prove that every group of order p^2 * q where p,q are primes is solvable...


    I've no idea about the second question. But the first one seems very easy, but I can't figure out how to solve it.


    Help is needed

    Thanks in advance
    This is for 2.

    Lemma 1. Any group of order p^2 is abelian for a prime number p.
    This can be proven by using that any p-group has a non-trivial center (Use a class equation). If G/Z(G) is cyclic or {e}, then G is an abelian group.

    Let |G|=p^2q and H be a Sylow q-subgroup of G. Then |G/H| = p^2, which is abelian by Lemma 1. We remain to show that H is normal in G. Recall that the number of H (Sylow q-subgroup) in G is, denoted n_q, divides |G| and having a form kq + 1. Now consider each case where q > p (If p>q, then you can show that there is a unique Sylow p-subgroup in G. This argument is relatively easy. If p=q, then we know that every finite p-group is solvable).

    Case 1: n_q = 1.
    H is normal in G, so \{e\} \triangleleft H \triangleleft G.

    Case 2: n_q = p.
    Since q > p, it never arises.

    Case 3: n_q = p^2.
    In this case, there are p^2 Sylow q-subgroup of order q and there are q Sylow p-subgroup of order p^2. If we have a unique Sylow p-subgroup of G, then we have a subnormal series, and we are done. Assume we have at least 2 Sylow p-subgroup of order p^2 in G. Any two distinct subgroups of order q intersect in the identity. Thus there are p^2(q-1) = p^2q -p^2 distinct elements of order q in G. Then, there should be p^2 distinct elements of order 1 or p or p^2. By our assumption, we have at least 2 Sylow p-subgroup of order p^2. Thus this situation neither arises by counting argument.

    Thus n_q = 1 and \{e\} \triangleleft H \triangleleft G, which implies that G of order p^2q is solvable.
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  4. #4
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    Thanks a lot to both of you
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