How many bijective functions from a set to itself containing a cycle that contains only 3 elements are there on the symmetric group (S4)?
My approach:
I understand that the formula is for l cycles in SN (symmetric group) is given by $nCl (l-1)!$ According to Wikipedia, but how do you prove it , and under what conditions does it fail?
On
No, this is incorrect. You have found the total number of permutations on $\{1,2,3,4\}$ however several of these do not contain a cycle of order three, for example the permutation $(1~2)(3~4)$.
The error in your calculation is that you are counting the number of strings of length three that you can make in your set, but when interpreting them as cycles you have duplicates. For example you counted $(1~2~3)$ to be different than $(2~3~1)$ and $(3~1~2)$ but in reality these are all the same cycle.
To correct your count, pick which three elements appear in your cycle simultaneously. Then, fix the smallest element of them to the front of the cycle so that it appears in canonical form. Finally, choose whether or not the largest remaining appears immediately to the right of the smallest in the cycle or not.
This gives $\binom{4}{3}\cdot 2 = 8$ such cycles. They are $(1~2~3),(1~3~2), (1~2~4),(1~4~2),(1~3~4),(1~4~3),(2~3~4),(2~4~3)$.
Here's the overview, as I've understood from the other answers and your approach:
The first and second points gives us 4 ways to pick the one we leave out, and 6 ways to order the 3 we do have. The last point is, that any three distinct elements, have three orderings in which you are simply moving the start point. Demonstration:
$$\begin{matrix}{\hspace{8pt}1\\\nearrow\searrow\\2\leftarrow3}& {\hspace{8pt}3\\\nearrow\searrow\\1\leftarrow2}&{\hspace{8pt}2\\\nearrow\searrow\\3\leftarrow1}\\{\hspace{8pt}1\\\nearrow\searrow\\2\leftarrow4}& {\hspace{8pt}4\\\nearrow\searrow\\1\leftarrow2}&{\hspace{8pt}2\\\nearrow\searrow\\4\leftarrow1}\\{\hspace{8pt}1\\\nearrow\searrow\\4\leftarrow3}& {\hspace{8pt}3\\\nearrow\searrow\\1\leftarrow4}&{\hspace{8pt}4\\\nearrow\searrow\\3\leftarrow1}\\{\hspace{8pt}4\\\nearrow\searrow\\2\leftarrow3}& {\hspace{8pt}3\\\nearrow\searrow\\4\leftarrow2}&{\hspace{8pt}2\\\nearrow\searrow\\3\leftarrow4}\\{\hspace{8pt}1\\\nearrow\searrow\\3\leftarrow2}& {\hspace{8pt}2\\\nearrow\searrow\\1\leftarrow3}&{\hspace{8pt}3\\\nearrow\searrow\\2\leftarrow1}\\{\hspace{8pt}4\\\nearrow\searrow\\2\leftarrow1}& {\hspace{8pt}1\\\nearrow\searrow\\4\leftarrow2}&{\hspace{8pt}2\\\nearrow\searrow\\1\leftarrow4}\\{\hspace{8pt}1\\\nearrow\searrow\\3\leftarrow4}& {\hspace{8pt}4\\\nearrow\searrow\\1\leftarrow3}&{\hspace{8pt}3\\\nearrow\searrow\\4\leftarrow1}\\{\hspace{8pt}4\\\nearrow\searrow\\3\leftarrow2}& {\hspace{8pt}2\\\nearrow\searrow\\4\leftarrow3}&{\hspace{8pt}3\\\nearrow\searrow\\2\leftarrow4}\end{matrix}$$ Note how the three in each row, simply rotate the others preserving order. The only difference, is in the value on top, of each triangle of arrows (Technically called $K_3$ I believe).