I would like to know if my proof below is correct.
Problem If $\tau = (1,2)(3,4,5)$ determine whether there is a $n$-cycle $\sigma$ ($n\geq 5$) with $\tau = \sigma^k$ for some integer $k$.
Solution: Let $\sigma = (a_1, a_2,\ldots,a_n)$. We actually prove a more general result here, i.e., $\sigma^k$ has a cycle decomposition with $d=\gcd(k,n)$ cycles, each having the same length $l=\dfrac{n}{d}$.
Pick any element, say $a_1$. Consider $\sigma^k$ acting on $a_1$. We then have $\sigma^k(a_1) = a_{1+k \pmod n}$. Let us look at the elements obtained by repeated action of $\sigma^k$ on $a_1$. In general, we have $\sigma^{kz}(a_1) = a_{1+kz \pmod n}$. For this to be equal to $a_1$, we need $kz\pmod{n} = 0$. The smallest positive $z$ gives the length of cycle containing $a_1$. The smallest value of $z$ is then $l=\dfrac{n}d$.
Now consider the smallest $a_i$ not generated by $a_1$ under $\sigma^k$ and repeat the process. This process has to terminate since $n$ is finite. Hence, $\sigma^k$ has a cycle decomposition with $d=\gcd(k,n)$ cycles, each having the same length $l=\dfrac{n}{d}$.
In our case, $\tau$ has a cycle decomposition where one cycle has a cycle length $2$ and another cycle has cycle length $3$. Hence, there is no $n$-cycle $\sigma$ ($n\geq 5$) with $\tau = \sigma^k$ for some integer $k$.
Thanks
Best Answer
Your more general result is correct, but they is some more work to be done in order to prove it. You need to show the following:
If $\gcd(n,k)=1$, then $\sigma^k$ is also an $n-$cycle.
If $\gcd(n,k)=d$, then $\sigma^k$ is a product of $d$ $\frac{n}{d}-$cycles.
You have done most of it. Does organise better your proof.