Problem. Suppose we have five objects arranged in some order $A, B, C, D, E$. Can we find some combination of permutations swapping around $3$ elements that result in the first two (or any two elements) switching places while leaving all elements in the same position?
For example, starting with $A, B, C, D, E$, we can swap $A, B, C$ around as a right rotation to get $B, C, A, D, E$. Can we combine any number of $3$-cycle permutations to get to $B, A, C, D, E$?
This was a problem suggested by a friend who has worked through a couple configurations and thinks that there might be a combination that works somewhere. While I have tried my hand at proving that this isn't the case, I'm not the most confident that the solution is right, so any help would be greatly appreciated. Here is my attempt:
Proof. We look to find a sequence $p_i$ of $3$-cycle permutations such that the following holds true: $$ (1 \quad 2) = \prod_{i} p_i.$$ Our claim is that no such sequence exists. Consider the parity of both sides. The sign of the left-hand side is $-1$ because it is an odd permutation. For the sign of the right-hand side, we use the fact that $\operatorname{sgn}$ is multiplicative. We have that \begin{align*} \operatorname{sgn}{\bigl( (1 \quad 2) \bigr)} &= \operatorname{sgn}{\prod_{i} p_i} \\\\ \iff -1 &= \prod_{i} \operatorname{sgn}{p_i} .\end{align*} Notice that because each $p_i$ is a $3$-cycle, they are all even permutations, necessarily having a sign of $1$. Because no product of $1$'s will ever equal $-1$, we conclude that there cannot exist such a sequence of permutations, and thus you cannot form a $2$-cycle out of several $3$-cycles. $\tag*{$\blacksquare$}$
I'm not the greatest at proofs, so I'm unsure if I've made too hasty of an assumption somewhere. If it turns out that my proof is incorrect, what would be an example of such a sequence of permutations? Once again any help will be appreciated.