Find a $6$-cycle $C$ such that $C^3=(2 \; 9) (5 \; 13) (11 \; 12)$

Question

My initial problem was:

Given $\sigma$ = $(1 \; 7 \; 3 \; 15) (2 \; 9) (4 \; 10 \; 6 \; 8 \; 14) (5 \; 13) (11 \; 12)$. Find $\tau$ such that $\tau^3 = \sigma$.

I know that there are $2$ cases.

• First case is when $\tau = C_1 * C_2 * C_3 * C_4 * C_5$, where length of $C_1=C_2=C_3=2$, $C_4$ is of length $4$, $C_5$ is of length $5$. I managed for the first case to find $C_1 = (2 \;9)$, $C_2 = (5 \; 13)$, $C_3 = (11 \; 12)$, $C_4 = (1 \; 15 \; 3 \; 7)$ and $C_5 = (4 \; 6 \; 14 \; 10 \; 8)$.
• Second case is when $\tau =C* C_4 * C_5$, where $C_4$ and $C_5$ are the same as in the first case and $C$ is of length $6$, since $\sigma$ has $3$ transpositions. But I've got stuck at finding $C$.

So, my remaining problem is:

Find a $6$-cycle $C$ such that $C^3=(2 \; 9) (5 \; 13) (11 \; 12)$ (or if possible, find all such $C$s).

Answer 1

The cube of an arbitrary $6$-cycle $$\rho=(a\;b\;c\;d\;e\;f)$$ is $$\rho^3=(a\;d)(b\;e)(c\;f).$$ The solutions $C$ can therefore be found the following way, after deciding which of your $6$ cyclically permuted numbers you will write first, i.e. to which number will $a$ correspond. Let us say for instance $a$ corresponds to $2$, hence $d$ must correspond to $9$.

• choose to which of the two remaining (unordered) pairs $\{5,13\}$ and $\{11,12\}$ will $(b\;e)$ correspond. $(c\;f)$ will then correspond to the other unordered pair of numbers.
• independently, order these two unordered pairs.

For instance if the first step was $(b\;e)\mapsto\{5,13\}$ (hence $(c\;f)\mapsto\{11,12\}$), and if you choose to order the two pairs $\{5,13\}$ and $\{11,12\}$ by respectively $5$ first and $11$ first, the correspondance will be $a\mapsto2,d\mapsto9,b\mapsto5,e\mapsto13,c\mapsto11,f\mapsto12$ hence the solution will be $$C=(2\;5\;11\;9\;13\;12).$$ The $7$ other solutions are $$\begin{matrix}&(2\;5\;12\;9\;13\;11)&(2\;13\;11\;9\;5\;12)&(2\;13\;12\;9\;5\;11)\\(2\;11\;5\;9\;12\;13)&(2\;11\;13\;9\;12\;5)&(2\;12\;5\;9\;11\;13)&(2\;12\;13\;9\;11\;5).\end{matrix}$$

Answer 2

You can do this individually for each of the disjoint cycles making up $\sigma$, and multiply the answers together.

Let $\tau$ be one of these cycles, of cycle length $n$. Note that none of the $n$ are divisible by 3, so there are integers $a,b$ such that $an + 3b = 1$. Then $(\tau^b)^3 = \tau^{1 - an} = \tau$, so $\tau^b$ is the third root of $\tau$ you are looking for.