I was working on the following problem:

Let $\sigma$ be the m-cycle $(1 2...m)$. Show that $\sigma^{i}$ is also an m-cycle iff $\gcd(i,m)=1$

A solution to this problem is given here.

But the solution I came up with doesn't seem as rigorous as the one given, so I wasn't sure if it is correct.

My attempted solution:

Suppose that $\operatorname{gcd}(i,m)=n$ where $1<n < m$.

Then we have:

$\sigma^{i}(a_{x_{1}})=a_{x_{2}}, \sigma^{i}(a_{x_{2}})=a_{x_{3}},...,\sigma^{i}(a_{x_{n}})=a_{x_{1}}$ where each of $a_{x_{j}}$ is an arbitrary element of $[1,2,3,4,....m]$

And since $n<m$, we have formed an n-cycle.

Thus $\operatorname{gcd}(i,m)=1$.

Now if $\operatorname{gcd}(i,m)=1$, then we have,

$\sigma^{i}(a_{x_{1}})=a_{x_{2}}, \sigma^{i}(a_{x_{2}})=a_{x_{3}},...,\sigma^{i}(a_{x_{m}})=a_{x_{1}}$ which forms an m-cycle.

Is this attempt correct?

Also if possible, can you give a shorter solution than the one given in the link above.

  • $\begingroup$ added "gcd" throughout your post, please double check that I haven't changed what you intended to write $\endgroup$ – angryavian Jan 3 '16 at 19:42

Personally I believe that you should add an explanation on why when $\text{gcd}(i,m)=n$ then you have $\sigma^i(a_{x_n})=a_{x_1}$. A similar problem can found in the case $\text{gcd}(i,m)=1$, you should add the details.

Then you should probably explain why this implies that $\sigma^i$ is not an $m$-cycle, of course one could say that from that it follows that $\sigma^i$ factors through an $n$-cycle, with $n < m$, so it cannot be an $m$-cycle.

It could be possible to provide a shorter proof but it would probably hide much of the lemmas of the proof in the link you gave.... so it could sound a little bit like cheating.


The ideas in your attempt are good, but not well explained. The second part is insufficiently motivated.

You have $$ \sigma^1(1)=2,\quad \sigma^2(1)=3,\quad \dots $$ (assuming $m\ge3$, of course). In general, $$ \sigma^i(k)=k+(i\bmod m) $$ as you can prove by an easy induction, where $i\bmod m$ denotes the remainder of the division of $i$ by $m$, with $0\le i\bmod m<m$.

Let $d=\gcd(i,m)$ and start from $1$. The first time you get back to $1$, successively applying $\sigma^i$, is after $m/d$ steps.


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