How many solutions the equation $x^3=(1\space 2\space 3\space 4)$ have in $S_7$ First I find the order of $(1\space 2\space 3\space 4)$,  which is $4$. We know $S_7$ has elements of order $12$. Hence $x^{12}=e$. So it has solution.  
I am unable to find the number of solutions. I think if I find the number of elements of order $12$ in $S_7$ then it would be the answer.
The answer is at least $2$, whether $S_7$ has $420$ elements of order $12$.
Please help me here to find the number of solutions.
 A: Hint: consider the cycle type of $x$.
A: For simplicity, denote $[n]=\{1,2,\cdots,n\}$.

Lemma 1: Let $S$ be an infinite set and $\sigma:S \to S$ be a finitary
  permutation -- a permutation that moves only finitely many elements.
  Then, a cycle decomposition for $\sigma$ is an expression of $\sigma$
  as a product of disjoint cycles.

First, due to Lemma 1, we can write $x$ as the composition of cycles, namely,
$$x=\sigma_1 \sigma_2\cdots\sigma_k,$$
where $\sigma_i$ are dijoint cycles with each other.
Without loss of generality,  We assert that $x=(1234)\sigma$ are the only solutions to $x^3=(1234)$, where $\sigma^3=(5)(6)(7)$.


*

*It is easy to verify that these are solutions to $x^3=(1234)$. To find out all such $\sigma$., note that the orders of elements in $S_3$ are 1, 2 and 3. Thus only those elements of order 1 or 3 satsify the requirement. Hence the $\sigma$ are as follows:$(5)(6)(7)$,$(567)$ and $(576)$. Consequently, the solutions are 
$$(1234)(5)(6)(7),(1234)(567),(1234)(576).$$

*To show that these solutions are the "only" ones, assume that
$(1234)$ was splitted into parts. Let $T_4=\{\sigma_i\mid \sigma_i
   \cap [4] \ne \emptyset\}$, where $\sigma_i$ was treated as a set.
If $\sigma_i \subsetneq [4]$, then $\sigma_i^3=I$ or $\sigma_i^3=\sigma_i$, where $I$ denotes the identity map. It is a contradiction by Lemma 1.
If $\sigma_i \not\subset [4]$, then there exists an element $a\in    [7]\setminus[4]$ such that $a\in\sigma_i$. Then $\sigma^3=(1234)$    would contain an element in $[4]$ and an element not in $[4]$. It is    also a contradiction by Lemma 1.
Note that according to Lemma 1, $x^3=\sigma_1^3 \sigma_2^3 \cdots \sigma_k^3$.
A: As $x^3$ has order $4$, the element $x$ must have order $4n$ for some number $n$. This means that $x$ must contain a $4$-cycle*, so $x$ has the form $(abcd)$, or $(abcd)(ef)$, or $(abcd)(efg)$. As $x^3=(1234)$, it follows that $x$ has the form $(abcd)$ or $(abcd)(efg)$.
It is then a simple exercise to play around with the numbers to work out what $(abcd)$ must be (note that $a, b, c, d\in\{1, 2, 3, 4\}$). There are then three solutions: $(abcd)$, $(abcd)(567)$, and $(abcd)(576)$. (You should convince yourself that there are not any more.)

*This is using the fact that $4=2^2$, that $4$ cannot be decomposed non-trivially into coprime factors. On the other hand, $(12)(345)$ has order $6$ because the length of the cyclces ($2$ and $3$ respectively) are coprime.
