Find all solutions in an equation with permutations in $S_{10}$

Question

Let $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 9 & 5 & 7 & 10 & 3 & 4 & 6 & 1 & 8\end{pmatrix} \in S_{10}.$

Find all permutations $\tau \in S_{10}$ where $\tau^3 = \sigma.$

My first intuition was to multiply the equation with $\sigma^{-1}$ so it would look like this: $\tau^3 \sigma^{-1} = \sigma\sigma^{-1}$, which would result in $\tau^3 \sigma^{-1} =e$.

I am not sure how to advance from here. I know how to calculate the order of $\sigma$, its number of inversions and the signature, but I don't know how I would use that information towards solving this problem.

Answer 1

The presence of a single $3$-cycle in $\sigma=(1,2,9)(4,7)(3,5,10,8,6)$ kills all hope for a solution to $\tau^3=\sigma$.

• We see that $\operatorname{ord}(\sigma)=\operatorname{lcm}\{3,5,2\}=30$.
• Recall the basic fact (from the theory of cyclic groups): if $c$ has order $n$, then $c^k$ has order $n/\gcd(n,k)$. So if $\tau$ has order $m$, then we must have $m/\gcd(m,3)=30$ implying $m=90$.
• But there are no element of order $90$ in $S_{10}$. The smallest symmetric group with elements of order $90$ is $S_{16}$ where there is room for a permutation of cycle type $(9,5,2)$.
• A different argument woud be to observe that $\tau^{10}$ would have order $9$, meaning that $\tau^{10}$ must be a $9$-cycle. This would imply that $\tau^{30}$ is a product of three disjoint $3$-cycles. But, $\tau^3=\sigma$ implies that $\tau^{30}=\sigma^{10}=(129)$, a single $3$-cycle. This is a contradiction.

The conclusion is that the cube of a permutation $\tau\in S_n$, no matter what $n$ is, cannot have a single $3$-cycle in its cycle decomposition. The analogous result holds for all primes $p$: the $p$th power of a permutation cannot have a single $p$-cycle (the number of $p$-cycles in a $p$th power must be a multiple of $p$).

Answer 2

$\tau^3=\sigma \implies \tau^{90}=e$, since $|\sigma|=30$. So $|\tau|\mid 90$.

It follows that the cycle decomposition of $\tau$ can consist only in cycles of length (order) dividing $90$; hence of length $1,2,3,5,6,9$ or $10$.

None of these gives a $3$-cycle when cubed. The cube of a $2$ cycle is another $2$ cycle. The cube of a $3$ cycle is $e$. The cube of a $5$ cycle is another $5$ cycle. The cube of a $6$ cycle is a product of three $2$ cycles. The cube of a $9$-cycle is the product of three $3$-cycles. Finally, the cube of a $10$ cycle is another $10$ cycle.

But $\sigma =(129)(351086)(47)$.

So there are no solutions.

Answer 3

Here is a much more pedestrian approach:

Suppose there is some $r$ such that $r^3 = \sigma$.

Note that $\sigma$ can be written uniquely (modulo order) as $\sigma = a b c$, where $a,b,c$ are disjoint cycles of length $2,3,5$ respectively. In particular, $\sigma$ contains a cycle of length $3$.

Let $r=d_1...d_m$, where $d_k$ are disjoint cycles. Hence $r^3 = d_1^3 ... d_m^3$

Some work shows that if $d$ is a cycle of length $l(d)$, then $d^3$ will have (possibly multiple) cycles of length:

\begin{array}{|c|c|c|c|} \hline l(d)& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline l(d^3) & 1 & 2 & 1 & 4 & 5 & 2 & 7 & 8 & 3 & 10 \\ \hline \end{array}

Hence one of the $d_i$ must have length $9$ and hence the others must be trivial, but this is a contradiction.

Hence there is no such $r$.