Show $(abcde)$ is the cube of a $5$-cycle.

Question

Show $(abcde)$ is the cube of a $5$-cycle.

Let $x^3= (abcde).$

$$x^3 = \begin{pmatrix} a & b & c & d & e \\ b & c & d & e & a\end{pmatrix} = (abcde)$$

Taking inverse, get:

$$x^{-3} = \begin{pmatrix} b & c & d & e & a \\ a & b & c & d & e\end{pmatrix} = (baedc)$$

There seems no way to take out $x$ from either $x^3$ or $x^{-3}$?

Answer 1

$(abcde)=(acebd)^3$

To get $\sigma^n$, where $\sigma$ is a cycle, just "jump $n$-times".

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Or, using the hint, you just need $(abcde)^2$.

Answer 2

Consider a $5$-cycle $x=(x_1\ x_2\ x_3\ x_4\ x_5)$. We want $x^3=(a\ b\ c\ d\ e)$, that is, $$x^3=(x_1\ x_2\ x_3\ x_4\ x_5)^3=(x_1\ x_4\ x_2\ x_5\ x_3)=(a\ b\ c\ d\ e).$$ One solution is $x_1=a$, $\ x_2=c$, $\ x_3=e$, $\ x_4=b$, $\ x_5=d$,
so $x=(a\ c\ e\ b\ d)$.