I am reading the book Dummit and Foote - Abstract Algebra . One of the exercises is to find an $n$-cycle $(n \ge 5)$, $\sigma$ such that $\sigma^k = \tau$ for some positive integer $k$, where
$$\tau=(1,2,3)(4,5).$$
My Try:
1) $n$ cannot be greater than $5$ because all elements of an $n$-cycle will have same order but if it is greater then some have order which is actually at most $\frac{n}{2}$. So $n=5$.
2) If there existed $5$-cycle then all elements will have order $5$. But $1,2,3$ have order $3k$ and $4,5$ have order $2k$. This is a contradiction. So there does not exist an $n$-cycle $(n \ge 5)$. Is my reasoning right? If not, please tell where I am going wrong. Thanks.
The key is a problem on the same page in Dummit and Foote which says that if $\sigma=(a_1,a_2,\cdots,a_n)$ is an $n$-cycle, then $\sigma^r$ carries each $a_k$ to $a_{k+r},$ where if the subscript $k+r$ happens to exceed $n$ it is reduced mod $n$.
Suppose then that under some $\sigma^r$ an element $a_i$ is carried to $a_j$, and $a_j$ is carried to $a_k$, and finally $a_k$ is carried back to $a_i$ (by the map $\sigma^r$). This means that the three cycle $(a_i,a_j,a_k)$ is one of the cycles in the cycle decomposition of $\sigma^r.$ But then by the remark above from the exercise, each other cycle in the decomposition will be a three-cycle, since for example it follows that $$a_{i+1} \to a_{j+1} \to a_{k+1} \to a_{i+1},$$ with the subscripts interpreted mod $n$.
In summary, the other exercise in Foote implies that a power of a single cycle has a decomposition consisting of cycles all of the same length. In particular one cannot get the example you are working on, since it is a product of a three cycle by a two cycle.