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Let $G$ be a cyclic group of order $n$. Suppose $x$, $y$ are two elements of order $d$, where $d$ divides $n$. Show that $y = x^m$, where $m$ is an integer coprime to $n$.

I know $y=x^m$ since the subgroups generated by $x$ and $y$ must be identical. I do not know how to show the coprimeness, however.

Johnny Apple
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2 Answers2

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This seems surprisingly tricky for such an elementary problem. For a positive integer $m$, let $U(m) = (\mathbb{Z}/m\mathbb{Z})^{\times}$ be the group of units modulo $m$. Then for any positive integers $d \mid n$, the quotient map

$\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/d\mathbb{Z}$ induces a map on unit groups

$U(n) \rightarrow U(d)$, which I claim is always surjective. Thus, if you start with something which is a unit modulo $d$, then you can always correct by a multiple of $d$ to get something which is a unit modulo $n$: that's what you're trying to prove.

How do you prove this fact? Carefully! It is enough to go from any $d$ to $dp$ for a prime $p$, and you want to treat the cases $p \mid d$ and $\operatorname{gcd}(p,d) = 1$ separately. I'll bet this question has been asked and answered on this site before, but in case not and you need more help, please ask.

Added: Indeed the surjectivity question has been asked here before: see this. The answer still leaves something to the reader, so still please feel free to ask for more help...

Community
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Pete L. Clark
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  • I don't understand how what you are asking me to do is equivalent to the problem, sadly. I'm guessing $y$ here is our $d$ unit, and would like to express it as an $n$ unit...but how does raising it to an integral power represent correction by a prime multiple of d? – Johnny Apple Aug 23 '13 at 05:26
  • @Anthony: Instead of working in "a cyclic group of order $n$", by choosing a generator one may as well work in $\mathbb{Z}/n\mathbb{Z}$. This converts to "additive notation":, i.e., $x^d$ becomes $dx$. It is now easy to see that if $x = my$ and $x$ and $y$ both have some order $d$, then $m$ must be coprime to $d$. Note that $m$ is only unique modulo $d$: since $y$ has order $d$ also $(m+ad)y = my + a(dy) = x + 0 = x$. Thus we want to correct by a multiple of $d$ to make the order prime to $n$. – Pete L. Clark Aug 23 '13 at 12:55
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Edit: This is nonsense.

Well, the coprime part is false. Here's a counter-example:

Take $\mathbb{Z}/n\mathbb{Z}$, where $n=12$, $x=3$, and $y=9$. Then both $x$ and $y$ are of order $4$, but $y=3x$. So we've found an $m$ that's not coprime to $n$.

On the other hand, if you change the prompt to "$m$ is an integer coprime to $d$," the statement becomes true (and therefore much easier to prove). :)

Vishal Gupta
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Josh Keneda
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  • The statement doesn't say that it must be true for every $m$, only that there is some $m$ for which it is true. Note that $m$ is only well-determined modulo $d$. In your case, also $y = 7x$. – Pete L. Clark Aug 23 '13 at 05:14