Exercise 1.3.12 (Is $(1\ 2)(3\ 4 \ 5)$ a power of a $n$-cycle?)

Proof.

(a)

Since $${(13579246810)}^5=(12)(34)(56)(78)(910),$$

there is a $10$-cycle $\sigma =(13579246810)$ such that $\tau ={\sigma }^5$.

(b)

Put $\tau =(12)(345)$.

Suppose for the sake of contradiction that there is some $n$-cycle $\sigma =(a_1{a}_2\dots a_n)$ such that $\tau ={\sigma }^k$ for some integer $k$, so that

$${\sigma }^k={(a_1{a}_2\dots a_n)}^k=(12)(345)=\tau .$$

Then ${\sigma }^{2k}(1)={\tau }^2(1)=1$. Since $\sigma$ is a $n$-cycle, $n\mid 2k$.

Similarly, ${\sigma }^{3k}(3)={\tau }^3(3)=3$. Therefore $n\mid 3k$.

From $n\mid 2k$ and $n\mid 3k$, we deduce that $n\mid 3k-2k$, so $n\mid k$ and $k=\mathit{qn}$ for some integer $q$. Therefore ${\sigma }^k={({\sigma }^n)}^q=e$. This gives $\tau ={\sigma }^k=e$, which is false. This contradiction shows that $\tau$ is not the power of any cycle.