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Math Help - Groups, Subgroups and Normal Subgroups

  1. #1
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    Groups, Subgroups and Normal Subgroups

    Hi there,

    I'm going over some past papers for an exam but with no answers and have come a cropper on this one, any help appreciated!

    Let H be a subgroup of a group G and let g \in G. Define

    H^g := \{ ghg^{-1} | h \in H \}

    i) Prove that H^g is a subgroup of G.

    ii) Let g, k, \in G. Which of these statements is true

    (H^g)^k = H^{gk} or (H^g)^k = H^{kg}

    Give a proof of the statement that is always true

    iii) Set N := \bigcap_{g \in G} H^g. Prove that N \triangleleft  G and that N is a subgroup of G that is contained in H.

    Pretty stuck except for ii) (H^g)^k = H^{kg} is obviously the right one (I hope!).
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  2. #2
    Member Sylvia104's Avatar
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    Re: Groups, subgroups and normal subgroups

    (i) and (ii) should be straightforward.

    Hint for (iii). Let G/H denote the set of all left cosets of H in G. Each element of g induces a permutation \pi_g of G/H via \pi_g(aH)=gaH for each left coset aH. Let \Pi=\left\{\pi_g:g\in G\right\}. The mapping f:G\to\Pi defined by f(g)=\pi_g is a homomorphism. Prove that N is the kernel of this homomorphism.
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  3. #3
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    Re: Groups, Subgroups and Normal Subgroups

    i) let x,y be in Hg.

    this means that x = ghg-1 for some h in H, and y = gh'g-1, for some h' in H.

    therefore, xy-1 = (ghg-1)(gh'g-1) = ghh'-1g-1,

    and hh'-1 is in H (since H is a subgroup), therefore xy-1 is in Hg, so Hg

    is a subgroup of G (by the one-step subgroup test).

    ii) note that (Hg)k = k(Hg)k-1 = k(gHg-1)k-1 = (kg)H(kg)-1 = Hkg

    iii) first, suppose that a,b are in N. this means that a,b are in every gHg-1, for any g in G. by part (i) above, each gHg-1

    is a subgroup, so ab-1 is in every gHg-1, so ab-1 lies in N, thus N is a subgroup.

    let x be any element of G, and let n be an element of N. note that N = ∩g Hg = ∩xg (Hg)x

    since as g runs over all elements of G, so does xg.

    but ∩xg (Hg)x = ( ∩g Hg)x = Nx, so N is normal.

    (to see that for two subgroups H,K, that (H∩K)g = Hg∩Kg, suppose that

    x is in (H∩K)g. then x = ghg-1, where h is in H∩K. hence h is in H, so ghg-1 is in Hg.

    but h is in K as well, so ghg-1 is in Kg. hence x = ghg-1 is in both Hg and Kg,

    so x is thus in their intersection. on the other hand suppose that x is in Hg and Kg. so

    x = ghg-1 = gkg-1, for some h in H, and k in K.

    if ghg-1 = gkg-1, then h = k, so h = k lies in H∩K, so x is in (H∩K)g

    the same proof holds for any arbitrary intersection of subgroups of G).
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