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Math Help - Two Problems In Group Theory

  1. #1
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    Two Problems In Group Theory

    Hello there

    Wow, this forum is amazing! I didn't realize places like this exist and I'm really looking forward to getting stuck in to helping people. Anyway, my first post is unfortunately pleading for help with a subject that I have very little in common with. However, I have an upcoming assessment and am really struggling. Any help that you could give me would be very much appreciated

    1/ Let G be a finite group and N be a normal subgroup of G such that gcd(|N|,|G/N|)=1. Show that N is the only subgroup of G with order |N|.
    I have the feeling that this has something to do with the Second Isomorphism Theorem, though I am not sure. I have tried everything I can think of involving it but to no avail

    2/ Give a decomposition series for the dihedral group of order 28. To which well-known groups are the composition factors isomorphic? I'm lost on this one

    Thank you in advance for any help that you can give

    Paul
    Last edited by notyeteuler; March 6th 2008 at 11:12 AM.
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  2. #2
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    Quote Originally Posted by notyeteuler View Post
    1/ Let G be a finite group and N be a normal subgroup of G such that gcd(|N|,|G/N|)=1. Show that N is the only subgroup of G with order |N|.
    I have the feeling that this has something to do with the Second Isomorphism Theorem, though I am not sure. I have tried everything I can think of involving it but to no avail
    I have to go soon. I did not solve it yet, but I can solve if for abelian groups, not sure if this is the approach you are looking for. Suppose that |G|>1 then |G| = p_1^{a_1}\cdots p_k^{a_k}. If H\subseteq G such that \gcd(|H|,|G/H|)=1 it means H=p_1^{b_1}\cdots p_k^{b_k} where b_i = a_i \mbox{ or }0. Suppose then H=p_1^{a_1}\cdots p_l^{a_l} let P_1,...,P_l be the p_1,...,p_l Sylow subgroups. Then P_1...P_l = H because there can be only one Sylow subgroup since the group is abelian all of them are normal. (What an ugly proof ).
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  3. #3
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    Thank you for your help However, we I have not yet learned about Sylow subgroups and expect that we are not meant to use them
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  4. #4
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    We can prove something stronger.

    Theorem: Let G be a finite group and N a normal subgroup with \gcd( |N|,|G/N|) = 1 and if H\subseteq G is a subgroup with (|H|) \big| (|N|) then H\subseteq N.

    Proof: Form the factor group G/N. Let a\in H and consider the coset aN, we will show that aN=N proving that a\in N and completing the proof. Note, it is sufficient to prove that the order of aN in the group G/N is 1. Let k be the order of aN. Let n be the order of a in the group G. Now (aN)^n = a^nN = N thus it means k|n. But k|(|G/N|) by Lagrange's theorem and n divides |H| so n divides |N| by Lagrange's theorem, thus k divides |N|. But then k=1 because they are relatively prime.
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  5. #5
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    Quote Originally Posted by notyeteuler View Post
    2/ Give a decomposition series for the dihedral group of order 28. To which well-known groups are the composition factors isomorphic? I'm lost on this one
    I assume you mean a composition series, there is no such thing as a decomposition series.

    The dihedral group \text{D}_{28} can be thought of \{ 1,x,x^2,...,x^{13},xy,x^2y,...,x^{13}y|x^{14}=y^2=  1 \mbox{ and }yx=x^{13}y\}.

    Let G_0 = \text{D}_{28}.
    Let G_1 = \left< x \right> = \{1,x,x^2,...,x^{13}\}.
    Let G_2 = \left< x^2 \right> = \{1,x^2,x^4,...,x^{12}\}.
    Let G_3 = \{1\}.

    Now G_3\triangleleft G_2\triangleleft G_1 \triangleleft G_0 is a composition series. Because G_0/G_1 \simeq \mathbb{Z}_2 which is simple. And G_1/G_2\simeq \mathbb{Z}_2 which is simple. And G_2/G_3\simeq \mathbb{Z}_7 which is simple.

    Note, all the composition factor groups are abelian which means this dihedral group is solvable. And furthermore, any composition series must has basically the composition series given above by the Jordan-Holder theorem (one of my favorite theorems from all math).
    Last edited by ThePerfectHacker; February 26th 2008 at 08:35 AM.
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  6. #6
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    Thank you very much, that's amazingly helpful
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