2
$\begingroup$

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.

  • 0
    So this is homework?2012-09-28
  • 0
    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

2 Answers 2

2

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.

  • 0
    I'm sorry but I'm very new to group theory and I don't understand what you're aiming at.2012-09-28
  • 0
    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.2012-09-28
  • 0
    So is the gcd(*n,s*) $\le$ gcd(*n,r*)? Because then $|\langle a^r \rangle|$ $\le$ $|\langle a^s \rangle|$?2012-09-28
  • 0
    I mean, am I on the right track with this line of thought? I know this isn't the condition.2012-09-28
  • 0
    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.2012-09-28
  • 0
    So I came up with this proof but it doesn't seem very solid. Maybe you have some suggestions:2012-09-28
  • 0
    Let gcd(*n,s*)=*e* and let gcd(*n,r*)=*d*. Assume *e*$|$*d*. Then *ep*= *d* for some *p*. Since $a^d$=$({a^e})^p$, we have by closure that $ \langle a^d \rangle$ $ \subseteq $ $ \langle a^e \rangle $. Which implies that $ \langle a^r \rangle$ $ \subseteq $ $ \langle a^s \rangle $.2012-09-28
  • 0
    Now assume that $ \langle a^r \rangle$ $ \subseteq $ $ \langle a^s \rangle $. Thus $ a^r $ $ \in $ $ \langle a^s \rangle $. In a finite cyclic group, the order of an element divides the order of the group, therefore $ |a^r| $ divides $ | \langle a^s \rangle | $. So, *q(n/d) = (n/e)*, for some *q*. Which implies that *eq = d*. So finally, *e* divides *d*.2012-09-28
  • 0
    Maybe I'll just submit it as an answer so it's easier to read.2012-09-28
  • 0
    Apparently I cannot answer my own question though. Ugh. Sorry. I must look like an idiot.2012-09-28
  • 0
    @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
2

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 $n = 10$: $$ \langle 2\rangle = \{0, 2, 4, 6, 8 \}. $$

  • 0
    So it seems as though my conjecture is correct, and s divides r. The proof of this property is giving trouble though.2012-09-28
  • 0
    $s$ divides $r$ modulo $n$. $\langle 2\rangle \subset \langle 8\rangle$ in the above example.2012-09-28