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Showing that a group isn't cyclic.

  1. Aug 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]\left( \mathbb{Z}/32\mathbb{Z}\right)^{*}[/tex] is not a cyclic group.

    2. Relevant equations

    3. The attempt at a solution

    A little calculator magic has showed that all elements in the group have order 8, but that doesn't seem like a very educational solution :). If anyone could explain why all elements have the same order, it would be much appreciated.
     
  2. jcsd
  3. Aug 11, 2007 #2

    matt grime

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    It cannot be that all of the elements have order 8 - the square of any element of order 8 has order 4. So, put the calculator away and think instead (or do it by hand).
    What is the order of Z/32Z*? Is there an element of that order?
     
  4. Aug 11, 2007 #3
    The order of (Z/32Z)* (the multiplicative group of units) is phi(32)=16. So if I can show that all elements have order 8 or lower, I've shown that none of them can be generators. I guess my problem is that I don't see how to do this without resorting to a calculator.

    I'm not sure what you mean by the square of any element of order 8 has order 4.
     
  5. Aug 11, 2007 #4
    Think about it, if a has order 8 then a^8=1, what about a^2 though? a^8=(a^2)^4 which says that a^2 has order 4, but besides that an even simpler reason that not all elements have order 8 is that the identity certainly doesn't have order 8.
     
  6. Aug 11, 2007 #5

    matt grime

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    If you don't see why the square of an element of order 8 has order 4, then you need to relearn your definitions of order.

    There is a unique cyclic group of order 16. You can count all the elements of all orders. You can then simply show that this group you have to play with fails to have the right properties.
     
  7. Aug 11, 2007 #6
    Ahh, I get it now. Thanks both of you.
     
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