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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.


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So this is homework? – Thomas Sep 28 '12 at 17:26
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. – mkeachie Sep 28 '12 at 17:31
up vote 2 down vote accepted

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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I'm sorry but I'm very new to group theory and I don't understand what you're aiming at. – mkeachie Sep 28 '12 at 18:26
Think of $a$ as being $1$, and of our group as additive, so $s$ is just $1+1+\cdots+1$ ($s$ times). Then $e$ is in a sense the simplest generator of the group generated by $s$. So it will turn out that one version of the condition will be that $e$ is a divisor of $d$. For example let $n=24$, $s=10$. Then the group generated by $s$ is just $0,2,4, 22$. If $r=8$ then subset condition holds, if $r=9$ it doesn't. – André Nicolas Sep 28 '12 at 18:38
So is the gcd(n,s) $\le$ gcd(n,r)? Because then $|\langle a^r \rangle|$ $\le$ $|\langle a^s \rangle|$? – mkeachie Sep 28 '12 at 19:15
I mean, am I on the right track with this line of thought? I know this isn't the condition. – mkeachie Sep 28 '12 at 19:15
Generally on the right track. You are right about size, but smaller and subset are different. It is $\gcd(n,s)$ divides $\gcd(n,r)$. Certainly we then have subset, that should be easy. – André Nicolas Sep 28 '12 at 20:16

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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So it seems as though my conjecture is correct, and s divides r. The proof of this property is giving trouble though. – mkeachie Sep 28 '12 at 17:41
$s$ divides $r$ modulo $n$. $\langle 2\rangle \subset \langle 8\rangle$ in the above example. – copper.hat Sep 28 '12 at 17:48

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