In $S_7$, find the permutation $x$ that satisfies the condition $x^2 = (1,2)(3,4)$. In $S_7$, $X^2 = (1,2)(3,4)$.
What is the permutation $X$ that satisfies the given condition?
 A: Hint: What would be the order of $X$?

As an alternative more hands-on approach, think what $X$ does to the element $1$ of the underlying set. After applying $X^2$ i.e. applying $X$ twice you get to $2$. After applying $X$ twice to $2$ you get to $1$. This is because $X^2$ contains the transposition $(1,2)$. So you have $$X: 1\to a\to 2\to b \to 1$$ where you don't know what $a$ and $b$ are.
What happens when you apply $X^2$ to $a$ and $b$? To $3$ and $4$? To $5, 6,$ and $7$?
A: Write $X$ as product of disjoint cycles. Then you see what happens by squaring. Cycles of even length become two new cycles of half the length and cycles of odd length keep their length.
$X^2$ has no cycles of odd length so $X$ should also only have cycles of even length. We actually need a cycle of length 4 whose square is $X^2$. The two possibilities are $(1324)$, $(1423)$. Now we can multiply this by a cycle of length two whose square is the identity. There are the following possibilities: $(56)$, $(57)$, $(67)$. So in total we get $2\cdot 3=6$ possibilities for $X$.
A: Edit: Since Mark Bennet asked to calculate the order as a hint, I'm hiding this part of my answer.

 Note that $|X^2|=2$. Also: $|X^2|=\dfrac{|X|}{(|X|,2)}$ where (|X|,2) is the greatest common divisor of $|X|$ and $2$ and |X| is the order of $X$. Thus $|X|=2(|X|,2)$ and so $|X|$ is an even number. Thus $((|X|,2))=2$ and so $|X|=4$.

Try to find a 4-cycle which satisfies your equality.
