In $S_7$, $x^2=(1234)$ has no solutions but $x^3=(1234)$ has at least two How would I go about showing that in $S_7$, the equation $x^2=(1234)$ has no solutions but the equation $x^3=(1234)$ has at least two solutions?
 A: This answer is not a theoretical approach but it is base on using GAP. By using the following codes in GAP environment; you can easily find the desired elements in second equation:

S_7:=SymmetricGroup(7);
e:=Elements(S_7);;
for i in [1..Size(e)] do if e[i]^3=(1,2,3,4) then Print(e[i],"\n"); fi; od;



(1,4,3,2)
(1,4,3,2)(5,6,7)
(1,4,3,2)(5,7,6)

A: You can show that the second equation has at least two solutions by finding examples.  
You can show that there is no $x$ satisfying the first equation by considering what the order of $x$ would be, and what the possible orders of elements of $S_7$ are.
A: You could approach the second question by considering what the orbit under $x$ of the point $1$ (the values obtained from it by repeatedly applying $x$) should look like. You know that $x^3$ sends $1$ successively to $2,3,4$ and then back to $1$ (that is $(x^3)^4(1)=1$) so you know the size of the orbit must divide $12$ (while there is not enough room for it to be equal to $12$), and that it must be at least $4$. In fact you can easily argue that $4$ must divide the size of the orbit (look what $x$ and $x^2$ do when applied to the subset $\{1,2,3,4\}$), so the orbit must be equal to $\{1,2,3,4\}$. So restricted to those elements $x$ must be a $4$-cycle, and restricted to the remaining elements $x^3$ must be the identity. Now you can determine all solutions for $x$.
A: Hint: You could consider the cycle structure of the elements in $\text{S}_7$. Which of them can make its square a 4-cycle, which of them can make its cube a 4-cycle?
