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A very simple cyclic group problem

  • Thread starter radou
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  • #1
radou
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Homework Statement



As the title suggests - this is very simple, I only want to check.

Let G be a cyclic group of order n. Then, for every integer k which divides n, there are elements in G of order k.

The Attempt at a Solution



Now, G = <a>, and a^n = e by definition. Let k be an integer which divides n, so we have n = kp, for some integer p. Then a^n = a^(kp) = (a^p)^k = e. Now, the only thing we need to check in order to prove that the order of a^p is k is that k must be the least positive integer such that the equality is satisfied (this is what I'm a bit unsure about, but I believe it's correct - I used this reasoning in another few problems). Assume there is an integer q < k such that (a^p)^q = e. But then we have a^(pq) = e, and pq is clearly less than kp = n, contradicting the fact that the order of a is n (i.e. that the order of G is n). Hence, a^p is an element of G of order k.
 

Answers and Replies

  • #2
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Completely correct!
 
  • #3
radou
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OK, thanks!
 

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