Let $(a_1a_2a_3a_4)$ and $(a_5a_6)$ be disjoint cycles in $S_{10}$. Show that there is no element x in $S_{10}$ such that $x^2=(a_1a_2a_3a_4)(a_5a_6)$ I came across this question in my self-study through Gallian's book with an explanation stating that since $|(a_1 a_2 a_3 a_4)(a_5 a_6)|=4$ then such an $x$ has order 8. With no further explanation as to why such an $x$ has order 8. I'm sure it's there in the text I'm just missing it, so I was hoping someone can explain why?
 A: Clearly $x^8=(x^2)^4=e$, and so the order divides $8$. But it also can't be any proper divisor of $8$. For then we would have $x^4=e$, and so $(x^2)^2=e$, contradicting the fact that $|x^2|=4$.
A: If $y=(x_1x_2x_3x_4)(x_5x_6)$ has order $4$ and if there exists an $x \in S_{10}$ such that $x^2=y$, then $x^8 = (x^2)^4=y^4=e$ and so the order of $x$ is either $8$ or a factor of $8$ i.e., $8$, $4$,$2$,$1$. So let us prove that the order of $x$ cannot be $1$,$2$, nor $4$.
If the order of $x$ were $1$, then $x$ would be $e$, and as $y$ is $x^2$, it would follow that $y$ would be $e$ as well, which is clearly not so. So the order of $x$ cannot be $1$.
If the order of $x$ were either $2$ or $4$, then the equation $x^4=e$ holds [make sure you see why]. This gives $x^4=(x^2)^2$ $=y^2=e$, which is also not true either. So the order of $x$ cannot be $2$ or $4$ either.
So then, the order of $x$ cannot be $1$,$2$, nor $4$ indeed, leaving $8$ as the remaining possibility for the order of $x$.
A: Basic fact about cyclic groups: $$\lvert x^k\rvert =n/(n,k),$$ where $\lvert x\rvert =n.$  (See here for proof.)
Therefore $4=\lvert x^2\rvert =n/(n,2).$
But then $4\mid n\implies (n,2)=2\implies n=8.$
