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Math Help - Cartesian product and cyclic groups

  1. #1
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    Cartesian product and cyclic groups

    A pressing question:

    If we are given two cyclic groups, and we take a cartesian product of the two, is the result also a cyclic group?
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  2. #2
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    \mathbb{Z}_2 \times \mathbb{Z}_2 is a group of order 4 where all it's elements have order 2
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    Excellent! Thank you very much
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    MHF Contributor Drexel28's Avatar
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    In fact, \mathbb{Z}_n\oplus\mathbb{Z}_m is cyclic iff (m,n)=1. To think of why this is we need some element of \mathbb{Z}\oplus\mathbb{Z}_m to be of exactly order mn but it is clear the order of \mathbb{Z}_n\oplus\mathbb{Z}_m is \text{lcm}(m,n). Therefore lastly noting that \text{lcm}(m,n)=\frac{|mn|}{(m,n)} we conclude that (m,n)=1. And since all cyclic groups of order \ell are isomorphic to \mathbb{Z}_{\ell} this is the case for any cyclic groups.
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    Quote Originally Posted by Drexel28 View Post
    In fact, \mathbb{Z}_n\oplus\mathbb{Z}_m is cyclic iff (m,n)=1. To think of why this is we need some element of \mathbb{Z}\oplus\mathbb{Z}_m to be of exactly order mn but it is clear the order of \mathbb{Z}_n\oplus\mathbb{Z}_m is \text{lcm}(m,n). Therefore lastly noting that \text{lcm}(m,n)=\frac{|mn|}{(m,n)} we conclude that (m,n)=1. And since all cyclic groups of order \ell are isomorphic to \mathbb{Z}_{\ell} this is the case for any cyclic groups.
    thank you for further elucidating the problem. We had recently learned about this part: " \text{lcm}(m,n)=\frac{|mn|}{(m,n)} we conclude that (m,n)=1" So it was pretty clear from the previous post with Z2 x Z2, but your explanation was definitely something that went into my notes. Thank you!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flabbergastedman View Post
    thank you for further elucidating the problem. We had recently learned about this part: " \text{lcm}(m,n)=\frac{|mn|}{(m,n)} we conclude that (m,n)=1" So it was pretty clear from the previous post with Z2 x Z2, but your explanation was definitely something that went into my notes. Thank you!
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