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My professor gave us this problem, wondering if anyone could help me out:

Suppose a is an element of order n in a group G. Find a necessary and sufficient condition for which $\langle a^r\rangle \subseteq \langle a^s\rangle$. Prove your assertion.

Thanks.

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    Yes. I figured immediately that s must be less than or equal to r. But I'm pretty confident that s must divide r as well. I'm just having difficulty proving that both ways.2012-09-28

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Hint: Without loss of generality we may assume that our group is the integers $0$ to $n-1$ under additon modulo $n$.

Let $d=\gcd(r,n)$ and $e=\gcd(s,n)$. Find a relationship between $d$ and $e$ that is equivalent to the given condition.

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    @NeilReed: One can answer one's own question, and even choose that as the answer to accept. Don't know why there was$a$problem. Typing math in comments is not pleasant. Apart from the fact that I have trouble reading, looks to me as if you have given a detailed and correct solution. Might be good if you *could* type it as an answer and get full feedback.2012-09-28
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Hint: Think about the cyclic group $G = \mathbb{Z} / n\mathbb{Z} = \langle 1 \rangle$ (under addition) for various natural numbers $n$. How does the (cyclic) subgroups look like? Answer: This look like this: $\langle s\rangle$ for an integer $s$. Now try to write down the elements of $\langle r \rangle$ and $\langle s\rangle$ for various values of $r$ and $s$. For example with $n = 10$: $ \langle 2\rangle = \{0, 2, 4, 6, 8 \}. $

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    $s$ divides $r$ modulo $n$. $\langle 2\rangle \subset \langle 8\rangle$ in the above example.2012-09-28