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Showing that a subgroup is subnormal in the original group G

  1. Aug 2, 2007 #1
    Question: Let [itex]G[/itex] be a group of order [itex]p^n > 1[/itex] where [itex]p[/itex] is prime. If [itex]H[/itex] is a subgroup of [itex]G[/itex], show that it is subnormal in [itex]G[/itex]. That is, I need to show that there is a chain of subgroups [itex]H=H_0 \triangleleft H_1 \triangleleft ...\triangleleft H_m = G[/itex], where [itex]m\leq n[/itex].


    Analysis: We can easily show by induction on n that there is a series of normal subgroups of G
    [itex]1=G_0 < G_1 < ... < G_n = G [/itex] such that [itex][G_{i+1}:G_i]=p[/itex].

    Since H is a subgroup of G, [itex]|H|=p^k [/itex] where [itex]k \leq n[/itex]. Since H is a p-group, H also has a series of normal subgroups of H.


    Perhaps if H is normal in G, then we can consider G/H which has a series of normal subgroups
    [itex]\bar{1}=\bar{H_0}< \bar{H_1} < ... <\bar{H_m}=G/H[/itex].
    By Fourth Isomorphism Theorem, does that mean we have a series of normal subgroups [itex]H=H_0 < H_1 < ...< H_m = G[/itex], where [itex]m\leq n[/itex]?



    Firstly, how do we know that H is normal in G? :uhh:
    Secondly, is [itex]H_i [/itex] normal in [itex]H_{i+1}[/itex]?

    Thanks for your time!
     
    Last edited: Aug 2, 2007
  2. jcsd
  3. Aug 2, 2007 #2

    mathwonk

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    i guess you want to show that the normalizer of H in G, is larger than H. that smells to me like the proof that a group of order p^2 is always abelian?
     
  4. Aug 2, 2007 #3

    mathwonk

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    well i think i got it. look at the center of G. lemma: the center is non trivial.

    then there are two cases, either H contains the center, use induction on G/H, or it does not, and then N(H) contains more than H.

    does that work? i.e. the induction case?
     
  5. Aug 2, 2007 #4

    matt grime

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    H_i will have index p in H_{i+1}. For p groups it is elementary to show that this means it is normal - as ever these things are just down to partitioning sets into orbits.
     
  6. Aug 2, 2007 #5
    Hi Mathwonk, you are right. If H = G or 1, there's nothing to show. So assume that H is a proper nontrivial subgroup of G. Using the class equation, we know that the center of G is non-trivial.

    Case 1. If Z(G) is not contained in H, then H is properly contained in <H, Z(G)> which is contained inside the normalizer of H in G. So H is properly contained inside [itex]N_G(H)[/itex] and H is normal in [itex]N_G(H)[/itex]. But (can someone answer this question that) is it so obvious that after repeating finitely many times, we will eventually get that [itex]N_G...(N_G(N_G(H))) = G[/itex]? Or are we done by induction?

    Case 2. If Z(G) is contained in H, then consider H/Z(G). Since |H/Z(G)|< |H|, we use induction which says that [itex]\bar{H}=H/Z(G)[/itex] is properly contained in [itex]N_\bar{G} (\bar{H})[/itex]. By Lattice Isomorphism Theorem, we're done.


    And Matt Grime, don't we have to prove that given any H=H_i in G, a subgroup H_{i+1} containg H_i exists such that H_i is normal in H_{i+1}?

    Do you think we should begin by showing that H is contained in a maximal subgroup M of G (M is maximal subgroup of G if M is a proper subgroup and there does not exist proper subgroups K of G such that M < K < G)? Then show that [G:M] = p and M is normal in G (use Mathwonk's idea using normalizer of M in G). Then since |M|<|G|, use induction to finish constructing the chain?

    But how we do show that such maximal subgroup M of G exists such that M contains H?
     
    Last edited: Aug 3, 2007
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