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Math Help - a question about normal subgroups and index.

  1. #1
    Senior Member abhishekkgp's Avatar
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    a question about normal subgroups and index.

    given that N is a normal subgroup of G. also H is given to be a subgroup of G.
    also given that G is a finite group with gcd(|G:N|,|H|)=1.

    prove that H is a subgroup of N.
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    Quote Originally Posted by abhishekkgp View Post
    given that N is a normal subgroup of G. also H is given to be a subgroup of G.
    also given that G is a finite group with gcd(|G:N|,|H|)=1.

    prove that H is a subgroup of N.
    (1) (HN)/N=H/(H \cap N)

    Since N is a normal subgroup of G and H is a subgroup of G, HN is a subgroup of G. Thus, |(HN)/N|=|H/(H \cap N)| divides [G:N] by (1) and Lagrange's theorem. It follows that |H/(H \cap N)| divides both [G:N] and |H|. Since gcd(|G:N|,|H|)=1, we see that |H/(H \cap N)|=1. Thus, H \cap N=H. We conclude that H is a subgroup of N.
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    Senior Member abhishekkgp's Avatar
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    discussing the solution.

    so your solution uses the fact that if N is a normal subgroup and H is a subgroup then
    (H intersection N) is a normal subgroup of H??
    is that right?

    (why \cap not working?)
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    If |H/(H \cap N)|=1, then H \cap N = H. If H \cap N = H, then H is a subgroup of N.
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    Senior Member abhishekkgp's Avatar
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    discussion

    i got that part... i am talking about the very first line in your first reply.

    there you have written HN/N = H/(H intersection N).
    For the RHS of the above equation to be well defined its necessary that
    (H intersection N) is a normal subgroup of H. is this correct??
    thanks in advance..
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    It is the second isomorphism theorem. Yes, H intersection N is a normal subgroup of H. Define a surjective homomorphism f:H--->HN/N such that its kernel is (H intersection N). Now the kernel of f has to be the normal subgroup of H.
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    Senior Member abhishekkgp's Avatar
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    thanks a lot. that helped.
    also i found another solution which you may find intersesting and more algebraic.
    here it is:

    suppose that h is in H. and suppose [G:N] = k, |H| = m.

    since gcd(k,m) = 1, there are integers r,s with rk + sm = 1.

    it may be that H is trivial, in which case H < N.

    so suppose not. let h ≠ e be chosen arbitrarily in H.

    then Nh = (Nh)^(rk + sm) = (Nh)^(rk)(Nh)^(sm)

    = [(Nh)^k]^r[(Nh)^m]^s

    = (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N,

    so h must be in N, thus H < N.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Alternatively one can us the first isomorphism theorem. Note that since N\unlhd G we have that G/N is a group with order \left(G:N\right). Consider then the canonical projection \pi:G\to G/N. We see then that the restriction \pi_{\mid H}:H\to G/N is also a homomorphism but since \left(|H|,\left|G/N\right|\right)=1 it follows that \pi_{\mid H} is the trivial homomorphism 1:H\to G/N:h\mapsto N. In particular it follows that \displaystyle \pi_{\mid H}\left(H\right)=HN=N but this is true if and only if H\leqslant N.

    P.S. I implicitly used the FIT by using it to prove that if \phi\in\text{Hom}\left(G,G'\right) with |G|,|G'|<\infty then \left|\phi(G)\right| divides both |G| and |G'|.
    Last edited by Drexel28; January 2nd 2011 at 09:27 PM.
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  9. #9
    Senior Member abhishekkgp's Avatar
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    Re: a question about normal subgroups and index.

    Quote Originally Posted by Drexel28 View Post
    Alternatively one can us the first isomorphism theorem. Note that since N\unlhd G we have that G/N is a group with order \left(G:N\right). Consider then the canonical projection \pi:G\to G/N. We see then that the restriction \pi_{\mid H}:H\to G/N is also a homomorphism but since \left(|H|,\left|G/N\right|\right)=1 it follows that \pi_{\mid H} is the trivial homomorphism 1:H\to G/N:h\mapsto N. In particular it follows that \displaystyle \pi_{\mid H}\left(H\right)=HN=N but this is true if and only if H\leqslant N.

    P.S. I implicitly used the FIT by using it to prove that if \phi\in\text{Hom}\left(G,G'\right) with |G|,|G'|<\infty then \left|\phi(G)\right| divides both |G| and |G'|.
    i had asked this long back but somehow i didn't see it and i checked it out today. brilliant!

    i got another solution which is cool too:
    Suppose |G:N| = k, |H| = m.

    since gcd(k,m) = 1, there are integers r,s with rk + sm = 1
    If H is trivial then  H \leq N and we are done.

    so suppose H is not trivial. Let h \neq e be chosen arbitrarily in H.

    then Nh = (Nh)^{rk + sm}= (Nh)^{rk}(Nh)^{sm}

    \Rightarrow Nh= [(Nh)^k]^r[(Nh)^m]^s

    \Rightarrow Nh= (N^r)(N(h^m))^s = N(Ne)^s = N(N) = N

    so h must be in N, thus H < N.
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