f interchanges $\alpha$ and $\beta$ but fixes $\gamma$ and $\delta$

Question

$f=(\alpha \beta), \; g=(\beta \gamma), \; h=(\gamma \delta)$

Is there a way to solve this that avoids finding what the permutations in options are ?

Answer 1

We observe that each of the candidate permutations is the conjugate of a transposition. For instance, looking at permutation $A$ we have $$ f \circ g \circ h \circ g \circ f = (f \circ g) \circ h \circ (f \circ g)^{-1}. $$ (For the last equality, we are using the fact that a transposition is its own inverse as well as the shoe-sock theorem.) It is a general fact that, given $n$, if we conjugate a transposition by any element of the symmetric group $S_n$, then the result is a transposition. The upshot is that each of your candidate permutations is a transposition. Thus to verify if one of the candidate permutations is the transposition $(\alpha \delta)$ you need only determine the image of one element. This already saves a lot of work, but we can further save work by noticing that in each candidate permutation, it is always the case that $f$ appears only once (B and D) or $h$ appears only once (A and C).

In permutations $A$ and $C$, $\delta$ does not move until $h$ is applied, so you only need to determine what happens to $\delta$ from the application of $h$ onward. For instance, in permutation $A$, we have $$ \delta \overset{h}{\mapsto} \gamma \overset{g}{\mapsto} {\beta} \overset{f}{\mapsto} \alpha, $$ so permutation $A$ is indeed the transposition $(\alpha \delta)$.

Similarly, in permutations $B$ and $D$, $\alpha$ does not move until you apply $f$, so you can trace what happens to $\alpha$ from the application of $f$ onward.