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Math Help - Group Order a product of primes

  1. #1
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    Group Order a product of primes

    Hello!

    If there is a group G of order pqr

    (for distict primes p, q and r)

    with subgroups H and K of G,
    where the order of H is pq
    and the order of K is qr

    I am trying to show that the order of H \cap K must be q.

    -----

    Not sure where to run with this one?

    I thought maybe that an intersection of two subgroups may be a subgroup of either of those subgroups,
    ie, H \cap K is a subgroup of H, so the order of H \cap K must divide the order of H,
    and similarily,
    and also H \cap K is a subgroup of K,
    so the order of H \cap K must divide the order of K

    but I am not sure if that is valid or not (is the intersection of two subgroups (H and K) a subgroup itself of either of the two orginal subgroups (H or K)?

    Any help appreciated!
    Thank you!!
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Your working is right (intersecting two subgroups gives a subgroup - if you do not know this, it is a good exercise to prove it), but only goes so far. You have basically shown that either,

    |H\cap K|=q or |H \cap K|=1.

    It is therefore sufficient to prove that there exists some 1 \neq g \in H\cap K.

    Can you prove that such an element exists?
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  3. #3
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    I will try proving that! Perhaps by one of the subgroup tests! Nice result that I did not previously know.

    Back to the question at hand though - I can't seem to think of a way to show some there is a non-identity element in the intersection!

    I am trying to recall any theorem regarding the fact that both H and K have a common divisor in their orders, so is it the case that they have elements of the same order ( I have a feeling that this is only for Albelian groups, Cauchy's theorem)- and even then, that still doesn't mean that they have the same element!

    Thanks.
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  4. #4
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    Quote Originally Posted by matt.qmar View Post
    I will try proving that! Perhaps by one of the subgroup tests! Nice result that I did not previously know.

    Back to the question at hand though - I can't seem to think of a way to show some there is a non-identity element in the intersection!

    I am trying to recall any theorem regarding the fact that both H and K have a common divisor in their orders, so is it the case that they have elements of the same order ( I have a feeling that this is only for Albelian groups, Cauchy's theorem)- and even then, that still doesn't mean that they have the same element!

    Thanks.


    Try using that \displaystyle{|H\cap K|=\frac{|HK|}{|H||K|}} and assuming that the intersection is trivial.

    Tonio
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  5. #5
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    Hey, that works! thank you!

    My only question is, how do you know

    {|H\cap K|=\frac{|HK|}{|H||K|}}

    holds?

    is that true for all groups H, K and internal direct product HK? can it be simple proved?

    Thanks!!
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  6. #6
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    Quote Originally Posted by matt.qmar View Post
    Hey, that works! thank you!

    My only question is, how do you know

    {|H\cap K|=\frac{|HK|}{|H||K|}}

    holds?

    is that true for all groups H, K and internal direct product HK? can it be simple proved?

    Thanks!!

    It is true even for subsets H, K of a group G. Try first proving it by yourself, and then google it. It's a pretty standard result.

    Tonio
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  7. #7
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    Wait, isn't that fraction upside down? Should be

    |HK|=\frac{|H||K|}{|H\cap K|}.
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  8. #8
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    Quote Originally Posted by topspin1617 View Post
    Wait, isn't that fraction upside down? Should be

    |HK|=\frac{|H||K|}{|H\cap K|}.

    Of course...well noted!

    Tonio
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  9. #9
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    Thank you! It is a fairly obvious result, now that I see it that way.
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