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Math Help - Sylow Theorems/Simple Groups

  1. #1
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    Sylow Theorems/Simple Groups

    Hi everyone, I have a couple of problems:

    1. Find up to isomorphism, all abelian groups of order 32.

    2. Prove that no group of order 56 is simple. Hint: use the Sylow Theorems.

    3. Let G be a group of order 35: a) Prove that G has only one 5-Sylow and one 7-Sylow that is, G(5) and G(7) respectively. b) Deduce that G is cyclic.

    Thanks for the help.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    1. Find up to isomorphism, all abelian groups of order 32.
    By the fundamental theorem the problem is trivial:
    \mathbb{Z}_2^5 (means the direct product taken 5 times)
    \mathbb{Z}_{16}\times \mathbb{Z}_2
    \mathbb{Z}_4\times \mathbb{Z}_4 \times \mathbb{Z}_2
    \mathbb{Z}_8\times \mathbb{Z}_2\times \mathbb{Z}_2
    \mathbb{Z}_8\times \mathbb{Z}_4
    \mathbb{Z}_{32}
    2. Prove that no group of order 56 is simple. Hint: use the Sylow Theorems.
    Note 56 = 2^3 \cdot 7. There is a Sylow 2-subgroup. If k is the number of such Sylow subgroups then k\equiv 1 (\bmod 2) and k|7 so k=1. Since there is only one Sylow 2-subgroup it must remain invariant under conjugation, i.e. it is a normal subgroup. And hence this group cannot be simple.

    (By the way a famous conjecture of Burnside says no group of even order > 2 can be simple was proven in 1980's. So you can use it if you want complete overkill).

    3. Let G be a group of order 35: a) Prove that G has only one 5-Sylow and one 7-Sylow that is, G(5) and G(7) respectively. b) Deduce that G is cyclic.
    Using the same argument as about there is only one Sylow 5-subgroup P_1 and only one Sylow 7-subgroup P_2. Form the group product P_1P_2, this is a subgroup since P_1\triangleleft G and |P_1P_2|=35 because P_1\cap P_2=\{ e \} since \gcd (|P_1|,|P_2|)=1.
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  3. #3
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    I do not think I explain #3 in detail. Does it make sense? I did skip a lot of details.
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  4. #4
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    Thanks for the help so far! I kind of get what you are talking about. Maybe if you add some more details I will have a clearer understanding. How can you tell that the group is cyclic?
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  5. #5
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    First |P_1P_2||P_1\cap P_2|=|P_1||P_2| since |P_1\cap P_2|=1 it means |P_1P_2|=5\cdot 7=35 because \gcd(|P_1|,|P_2|)=1 (you understand that?) So P_1P_2=G and P_1\cap P_2=\{ e \}. Now there is a theorem is two normal subgroups have their intersection the identity and their group product equal to the whole group then the group is isomorphic to their direct product. Thus, G\simeq P_1\times P_2 \simeq \mathbb{Z}_5\times \mathbb{Z}_7\simeq \mathbb{Z}_{35}
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  6. #6
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    I'm following how to show that G is isomorphic to Z35, but how do you show that G is cyclic? You have to show that G is generated by something don't you?
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  7. #7
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    It is isomorphic to Z35 which is cyclic. So it is cyclic.
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  8. #8
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    ThePerfectHacker seems to have overlooked a possible scenario for your second question -- what if k=7?

    Let's start again, but consider Sylow 7-subgroups of a group G with order 56. By Sylow, there are n = 1 (mod 7) such subgroups, but also n divides 8 (by n_p the index in G of N_G(P), the normalizer of any Sylow p-subgroup). So the only possibilities are n=1, 8. If n=1, as ThePerfectHacker said, there is a unique Sylow 7-subgroup, which is normal, by conjugation preserving order of a subgroup.

    But what if n=8? Well, note that each Sylow 7-subgroup is isomorphic to Z_7 (as is any group of order 7). Crucially, each pair of distinct Sylow 7-subgroups is disjoint (setting aside the identity), because any common element would be a generator for both Sylow 7-subgroups. So there are 8*6=48 distinct, non-identity elements in the union of Sylow 7-subgroups.

    None of these 48 elements can be in a Sylow 2-subgroup (why?). That leaves 8 elements (including the identity) to form a single, unique Sylow 2-subgroup (we know at least one exists).
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  9. #9
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    Thank you for the correction
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  10. #10
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    No problem. It's easy to overlook the other case, unless of course you recently were forced to classify all groups of order 56 (many of which are generated by a semidirect product). Not the most fun task.
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