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


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.

  • $\begingroup$ What does "all elements of an $n$-cycle will have same order" mean? An $n$-cycle is a particular element of the symmetric group, so it has an order. What are its elements? $\endgroup$ – Sammy Black Apr 30 '14 at 21:29
  • 1
    $\begingroup$ Does the statement of the exercise say to actually find such a cycle, or does the problem just ask whether there exists such a cycle? $\endgroup$ – coffeemath Apr 30 '14 at 22:06
  • $\begingroup$ @coffeemath yes it asks whether it exists $\endgroup$ – happymath May 1 '14 at 5:19
  • $\begingroup$ @SammyBlack i mean $(a_i,a_{i+1},a_{i+2})$ is a cycle then all the elements of cycle have same order 3. $\endgroup$ – happymath May 1 '14 at 5:21

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.

  • $\begingroup$ Yes, you can "work backward" e.g.(132)(456) is the square of (143526). $\endgroup$ – coffeemath May 1 '14 at 5:23

The order of a permutation $\sigma \in S_n$ is the smallest positive power of $\sigma$ that equals the identity. In other words, $m = |\sigma|$ means that $$ \sigma^m = 1 \qquad \text{and} \qquad \sigma^i \ne 1 \text{ for all } 1 \le i < m. $$ Given an element $\sigma$ in the symmetric group (in any group, actually), you can consider the cyclic subgroup generated by $\sigma$: $$ \langle \sigma \rangle = \{ \sigma^i \mid i \in \Bbb{Z} \} = \{ 1, \sigma, \ldots, \sigma^{m - 1} \}, \text{ where } m = | \sigma |. $$

These facts will help you solve your problem:

Fact 1. Lagrange's Theorem.

Fact 2. For any $\sigma \in S_n$, $| \sigma^k |$ divides $| \sigma |$.

Fact 3. If $\sigma = (i_1, \ldots, i_p)(j_1, \ldots, j_q)$, a product of disjoint cycles, then $| \sigma | = \operatorname{lcm}(p, q)$.


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