What is an easy way to determine the elements of $S_{4}$?
While going through my revision process in an attempt to stem out the nitty gritty areas that I am unsure of, I chanced upon this. I tried to search for an answer via Google but there are no satisfying responses.
Any help is appreciated.
$\endgroup$ 33 Answers
$\begingroup$List all possible 4-letter words in the letters $1,2,3,4$ without repetition.
A systematic way is to list them in lexicographical order:
1234
1243
1324
1342
...
4321A word $a_1 a_2 a_3 a_4$ corresponds to the permutation $i \mapsto a_i$ in $S_4$.
$\endgroup$ 1 $\begingroup$Start with the identity (1)(2)(3)(4).
Now start by introducing some transpositions:
(1, 2)(3)(4), (1,2)(3,4), (1)(2,3)(4), (1,4)(2,3),
...
Then you do triples
(1,2,3)(4), (1)(2,3,4),
...
Finally (1,2,3,4)
If you think of the parenthesis as spacers and open a probability coursebook, there are some efficient ways to go about getting all 24
$\endgroup$ 0 $\begingroup$We know that every element of $S_4$ is an automorphism over ${1,2,3,4}$. We want to count the number of elements in $S_4$ then lets start with $1$, it must be mapped to an element in ${1,2,3,4}$ - so it has 4 options, now for 2 - he has only 3 options for mapping (since $1$ already occupied an element in ${1,2,3,4}$), in the same way $3$ has $2$ options, and the mapping of $4$ is determined uniquely after determining all the other mappings.
Therefore, we finally get $4\cdot 3\cdot 2\cdot 1 = 4! = 24$ such elements.
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