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Math Help - Non isomorphic groups

  1. #1
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    Non isomorphic groups

    Hi

    I need to find four non-isomorphic groups of order 2268 (expressed as products of their p-primary components), each isomorphic to A=Z126 x Z108.

    Any ideas? Do I start with the primary decomposition Z2^2 x Z3^4 x Z7
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  2. #2
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    Quote Originally Posted by moolimanj View Post
    Hi

    I need to find four non-isomorphic groups of order 2268 (expressed as products of their p-primary components), each isomorphic to A=Z126 x Z108.
    How can non-isomorphic groups be isomorphic to the same group!

    Any ideas? Do I start with the primary decomposition Z2^2 x Z3^4 x Z7
    You can use abelian groups. But you do not have to.
    The dihedral group D_{1134} is one.
    The cyclic group \mathbb{Z}_{2268} is another.
    This group S_3\times \mathbb{Z}_{378} is another.
    You can now use \mathbb{Z}_2^3 \times \mathbb{Z}_3^4 \times \mathbb{Z}_7.
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  3. #3
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    Thanks, but I dont understand your working. Perhaps I have interpreted the question wrong - can you explain a bit more. I have attached the original question:
    Attached Thumbnails Attached Thumbnails Non isomorphic groups-2268.jpg  
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  4. #4
    Super Member Matt Westwood's Avatar
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    Aha, that makes more sense.

    I presume that the A we're talking about is:

    <br />
A = \mathbb{Z}_{126} \times \mathbb{Z}_{108}<br />

    So right, the way I would attack this would be to factor 126 and 108 and work out the ways you can express 2268 in such a way as for it to be x \times y where x is a factor of 126 and y is a factor of 108.

    But it's a few years since I did M336, maybe someone else can come up with a fuller answer than this.
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  5. #5
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    Well, 2268 can be expressed in the following ways:

    Z2268
    Z2 X Z1134
    Z3 X Z756
    Z6 X Z378
    Z9 X Z252
    Z18 X Z126
    Z3 X Z3 X Z252
    Z3 X Z6 X Z126
    Z3 X Z3 X Z3 X Z84
    Z3 X Z3 X Z6 X Z42

    But ow do i identify the non-isomorphic groups? Whats the method
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  6. #6
    Super Member Matt Westwood's Avatar
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    Can't remember the details, but because Z126 and Z108 are not coprime, their product can not be a cyclic group. It's there in the notes somewhere.

    So if you factor Z126 and Z108 and work out the bits of it that *are* coprime, you can work out the maximum order of a cyclic group that A can have, and you'll probably find that 268 is not a factor of it.
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  7. #7
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    I still cant see the answer.

    for Z126 I get Z2 x Z3^2 x Z7
    and for Z108, I get Z2^2 x Z3^2

    Thus A = (Z2 x Z2^2) x (Z3^2 x Z3^3) x Z7

    Any ideas?
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  8. #8
    Super Member Matt Westwood's Avatar
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    Z2268 is a cyclic group. So by one of the results you'll have encountered already, all its subgroups are cyclic.

    Z126 = Z2 x Z3^2 x Z7 (all three of these are coprime).

    Z108 = Z2^2 x Z3^3 (both of these are coprime)

    To get Z2268 you need factors Z2^2 x Z3^4 x Z7.

    Z2268 needs a factor Z3^4 = Z81 and you can't get that by combining factors of Z126 and Z108, because, say, Z3 x Z3^3 is not a cyclic group. Zm x Zn is cyclic only when m and n are coprime.

    Study that part of your module again (see what I said last posting) and you'll find what I'm talking about.
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