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Math Help - Abstract Algebra- order of a group

  1. #1
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    Abstract Algebra- order of a group

    Compute |U(4)|, |U(10)|, |U(40)|

    I know that I have to compute the order of the group for each. Am I looking for |U(4)|=2, |U(10)|=4 and |U(40)|=10? as the answer?

    Thanks in advance
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mathatlast View Post
    Compute |U(4)|, |U(10)|, |U(40)|

    I know that I have to compute the order of the group for each. Am I looking for |U(4)|=2, |U(10)|=4 and |U(40)|=10? as the answer?

    Thanks in advance
    When ever you find the order of U\!\left(n\right) (same as \mathbb{Z}_n^{\times}), we are always guaranteed to have 1 and n-1 in U\!\left(n\right), so \left|U\!\left(n\right)\right|\geq2 when n>2.

    In your case,

    U\!\left(4\right)=\{1,3\}, so \left|U\!\left(4\right)\right|=2.

    U\!\left(10\right)=\{1,3,7,9\}, so \left|U\!\left(10\right)\right|=4.

    U\!\left(40\right)=\{1,3,7,9,11,13,17,19,21,23,27,  29,31,33,37,39\}, so \left|U\!\left(4\right)\right|=16.

    So two of your answers were correct.

    --------------------------------------------------------

    As an aside (if you see something like this in the future): If p is a prime, then \left|U\!\left(p\right)\right|=p-1
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  3. #3
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    Thank you, Chris
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    MHF Contributor Bruno J.'s Avatar
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    And, just to add : more generally,

    U(n)=U(\mathbb{Z}/n\mathbb{Z}) = \phi(n) = n \prod_{p\mid n}\left(1-\frac{1}{p}\right)
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