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Any group of order 952 contains a subgroup of order 68?

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I am struggling with a proof for this. Obviously Sylow's theorems come into play. We have that |G| = 952. As sylow's first theorem only covers subgroups of order pn, we cannot directly use it to assert the existence of a subgroup of order 68. On the other hand, if we can come up with a (normal) subgroup H of G such that [G : H] = 14, then by lagranges theorem we would have that |H| = 68. I am just unsure of how to begin with this construction.

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: May 9, 2012
  2. jcsd
  3. May 9, 2012 #2

    jgens

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    Start by using the Sylow Theorems to prove that there is a unique subgroup N of order 17.

    Edit: My first response gave too much of the solution away. The main idea was to think about internal semi-direct products.
     
    Last edited: May 9, 2012
  4. May 9, 2012 #3
    You can use Sylow's first theorem to show that a sylow p-subgroup of order 17 exists. Then sylow's third theorem shows that it is unique. Finally, Sylow's second theorem asserts that it is normal in G. Now...
     
    Last edited: May 9, 2012
  5. May 9, 2012 #4

    jgens

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    If we are thinking about constructing the subgroup of order 68 via an internal semi-direct product and we have a normal subgroup of order 17, then what kind of group should we be looking for now?
     
  6. May 9, 2012 #5
    I apologize jgens, I don't recall my abstract algebra class going into depth discussing semi-direct products. While I do think I could grasp them (after looking them up online) I don't want my instructor to think I simply copied a proof I don't understand. Is there an alternative proof strategy?
     
  7. May 9, 2012 #6

    jgens

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    Off the top of my head, I cannot think of an alternative way of proving this result without appealing to semi-direct products in some form. The problem is that most of the relevant theorems giving us information about the subgroups of G concern p-groups, while we want to find a subgroup which is not a p-group. The most obvious way of doing this is by taking two p-groups and combining them in some way to form another subgroup of G with the desired properties. And the easiest way of combining these groups is via semi-direct products.

    In any case, you can probably prove the following theorem without too much difficulty: If [itex]G[/itex] is a group, [itex]N,K \leq G[/itex] are subgroups of [itex]G[/itex] and [itex]N[/itex] is normal in [itex]G[/itex], then [itex]NK = \{nk \in G:n \in N, k \in K\}[/itex] is a subgroup of [itex]G[/itex]. The idea is to essentially use this result to build the group you want.
     
  8. May 10, 2012 #7
    I received this hint:

    The key result is the fact that if N is normal in G, then there is
    a 1-1 correspondence between subgroups of G containing N and subgroups
    of the factor group G/N. If H is a subgroup of G containing N, then
    the corresponding subgroup of G/N is H/N. So, you also have a simple
    relationship between the orders of the subgroups in the above 1-1
    correspondence (i.e. just multiply or divide by |N|). So, knowledge
    of subgroups in G/N can tell you about subgroups of G.

    Now..You can determine which subgroups of G are normal in G by Sylow's theorem. How do you determine the correct H containing N such that |H| = 68?
     
    Last edited: May 10, 2012
  9. May 10, 2012 #8

    jgens

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    Hmm ... I didn't think about using the lattice isomorphism theorem, but this proof is essentially the same as the proof I was suggesting (although the lattice isomorphism theorem takes care of the semi-direct product). In any case, you want to find a subgroup of G/N with order 4. Can you figure out why?
     
  10. May 10, 2012 #9
    Use the sylow theorems to show that the sylow 17-subgroup (call it K) is unique and hence normal in G. Hence, G/K is a group of order [G : K] = 56. But 56 = (22)(14), and so by Sylow's first theorem, G/K contains a subgroup of order 4.

    How should I integrate the lattice isomorphism theorem here? I know all groups of order 4 are abelian and hence normal, but I was unsure as to whether this is the right track.
     
  11. May 10, 2012 #10

    jgens

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    It's not quite that simple: Notice that 56 = 237 so Sylow's Theorem actually gives you a subgroup of order 8, not order 4. An abelian subgroup is also not necessarily normal (it is a good exercise to find an example of an abelian subgroup which is not normal).

    In any case, once you correctly find a subgroup of G/N or order 4, then just apply the lattice isomorphism theorem directly to get your group of order 68.
     
  12. May 10, 2012 #11
    I'm trying to figure out how to construct subgroups of factor groups
     
  13. May 10, 2012 #12

    jgens

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    Well start by using Sylow's Theorem to get a subgroup of G/N of order 8. Now for any group of order 8, can you find a subgroup of order 4?
     
  14. May 10, 2012 #13
    We've already shown above that a normal subgroup N of order 17 exists in G. Since G/N is then a group of order 56, we reapply sylow's theorems in a similar fashion to show that there exists a normal subgroup of G/N (which has the form L/N) having order 8. Applying Sylow's first theorem to L/N implies that it contains a subgroup K/N of order 22 = 4. By transitivity of subgroups, then, K/N ≤ G/N. Thus, by the lattice isomorphism theorem, there is a corresponding subgroup K of G of order [K : N]|N| = (4)(17) = 68.

    This seems more valid than previous attempts.
     
  15. May 10, 2012 #14

    jgens

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    If you want to talk about normality, then you should prove it. I don't know if there is always a normal subgroup of order 8, so it might not be possible. Luckily, it doesn't matter, so you don't need to mention it.

    I recommend you read through the Sylow theorems again. None of the Sylow theorems allow you to conclude this.
     
  16. May 10, 2012 #15
    My text has:

    Sylow's first theorem: Let pn divide |G|. Then G has a subgroup of order pn.
     
  17. May 10, 2012 #16
    I checked another book... it can't be the case that p divides m, where |G| = pnm. Strange that the first book states it that way then.
     
  18. May 10, 2012 #17

    jgens

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    The first theorem is usually stated as: Let G be a group and suppose that |G| = pnm where p does not divide m. Then G has a subgroup of order pn.

    If there is a theorem which says "If pn divides |G|, then G has a subgroup of order pn" this is a stronger statement than the first Sylow Theorem (at least as its usually stated). I am actually not convinced this stronger statement is true and I don't really have time to think through it. But if you can use that result, then the proof is fine.

    Edit: Does the first book state that G has a p-Sylow subgroup or Sylow p-subgroup? Because both of those terms mean that p cannot divide m.
     
  19. May 10, 2012 #18
    Every group of order 8 has a subgroup of order two (isomorphic to Z2). Perhaps we can form the direct product of this group of order 2 to achieve our group of order 4?
     
  20. May 10, 2012 #19

    jgens

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    You are working too hard. There are five groups of order 8: [itex]Z_8,Z_4 \times Z_2,Z_2 \times Z_2 \times Z_2,D_8,Q[/itex]. Each of these has a subgroup of order 4 (you should prove this).

    Edit: You should do this without appealing to the Sylow Theorems.
     
    Last edited: May 10, 2012
  21. May 11, 2012 #20
    Jgens, once I prove that all groups of order 8 have a subgroup of order 4, I can place this in the proof and use the lattice isomorphism theorem to show that a subgroup H of order 68 exists in G.


    Before I do this however, I figure I pose another question as to the validity of one of my proofs before it gets too late:

    Show that the extension field Q(√2,√3) of the rationals is equal to the simple extension Q(√2 + √3).

    Proof: Clearly Q(√2,√3) is a subset of Q(√2 + √3) as Q(√2,√3) contains both √2 and √3 and hence also √2 + √3 by the field operations. Now, for nonzero p, we can divide p(√2 + √3) by p to obtain elements of the form q + √2 + √3, where of course the field operations allow √2 and √3 to be multiplied by any element of Q. Note too that(√2 + √3)(√2 + √3) = 2 + 2√2√3 + 3 which is also clearly expressible in Q(√2,√3) as √6 = √2√3.
     
    Last edited: May 11, 2012
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