Let $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 5 & 7 & 1 & 4 & 6\end{pmatrix} \in S_{7}.$
How many solutions does $x^3 = \sigma^6$ have?
Here is what I have done so far:
If we decompose $\sigma$ into disjoint cycles we get $\sigma = (1 2 3 5)(4 7 6)$. From this remark we deduce that $ord(\sigma) = lcm(3, 4) = 12$.
Since $ord(\sigma) = 12$, this implies that $ord(\sigma^6) = \dfrac{12}{gcd(12, 6)} = 2$, which results in $ord(x^3) = 2.$
Suppose $m = ord(x)$. We can calculate $m$ from this equation: $ord(x^3) = \dfrac{m}{gcd(m, 3)} = 2.$
That equation implies $m = \{2, 6\}$ and elements of order $2$ and $6$ indeed exist in $S_{7}.$
So basically what I understand from this is that we have an unknown permutation, $x$, which has the order of either $2$ or $6$ which is equal to a permutation of order $2.$
Does this mean that $x$ has $2$ solutions? Sorry if it's obvious but I barely grasp the concepts of modern algebra.