3

I'm given a $12$ cycle $\sigma = (1\dots 12)$, and want to find for which $i$ is $\sigma^i$ also a $12$-cycle.

We know that if $G$ is a group and $g \in G$, with $|g| = n$, then $|g^a| = \frac{n}{(n,a)}$, so in this case I want to say that $i$ must be coprime with $12$.

However, how do we eliminate the scenario where we have a permutation of order 12 which isn't a 12-cycle? That is, how do we deal with say $\sigma' = (123)(4567)$ which isn't a $12$-cycle but does possess order $12$?

Thanks in advance for the clarifications.

Shaun
  • 47,747
Irving Rabin
  • 2,754
  • If $k$ is coprime to $n$, then ${0,k,2k,\cdots,(n-1)k}$ is an exact set of representatives for the integers mod $n$. If we take $(a_0~a_1~\cdots~a_{n-1})$ to the power of $k$ we get $(a_0~a_k~\cdots~a_{(n-1)k})$ which is thus still an $n$-cycle because the entries, as listed, are all distinct. – anon Oct 16 '21 at 18:12

2 Answers2

5

You have to show that every element of $\{1,\dots, 12\}$ is in the same cycle as $1$ in the permutation $\sigma^k$ when $k$ Is coprime to $12$. To do this you must show for each $a$ there is an $r$ so that $(\sigma^k)^r(1) = a$. This indeed happens because in general $\sigma^m(1) \equiv 1+m \bmod 12$. So you need $kr\equiv a-1 \bmod 12$ which is possible since $k$ is coprime to $12$.

Asinomás
  • 107,565
  • When you say that every element is in the same cycle as $1$ in $\sigma^k$, this means that $\sigma^k$ does not split into distinct cycles correct? Is there any way you could explain the condition that $(\sigma^k)^r(1)=a$? How does this demonstrate that each element is in the same cycle as $1$? – Irving Rabin Oct 16 '21 at 18:25
  • 1
    The elements in the cycle of $x$ in a permutation $\tau$ are those that are of the form $\tau^n(x)$ for some $n$. – Asinomás Oct 16 '21 at 18:26
  • 1
    Perfect, thank you. – Irving Rabin Oct 16 '21 at 18:35
3

This is obvious if you remember that if $\sigma=(a_1\,a_2\,\dots\,a_i\,\dots a_n)$ is an $n$-cycle, we have $$\sigma^k(a_i)=a_{(i+k)\bmod n},$$ and translations of the additive group $\mathbf Z/n\mathbf Z$ are bijections of the group.

Bernard
  • 179,256