So I'm trying to find and write down the left and right cosets of the subgroup H generated by (1 3 2 4) in $S_4$. I know that the elements of $S_4$ = {(e), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4), (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2), (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3)}, and the subgroup generated by (1324) is H = {e, (1,3,2,4), (1,2)(3,4), (1,4,2,3)}, where e is the identity, and I think that the group generated by (1 3 2 4) is H = {e, (1 3 2 4), (1 2)(3 4), (1 4 2 3)}.
From Lagrange, we know that there will be 24/4 = 6 cosets (I think there should be 12, 6 left and 6 right?). The left coset generated by (1 2) is $L_{1,2}$ = {(1 2), (1 3)(2 4), (3 4), (1 4)(2 3)}. I'm stuck on the second left coset that I tried to write down, $L_{1,3}$. I have (1 3), (1 2 4 3), and (1 2 3 4) as elements of the coset, but I tried to multiply (1 3)(1 4 2 3) and got (1 4 2 3).
So, my two questions: I'm pretty sure the last element is fine, but I tried checking it by using the WolframAlpha "permutation" function and got something that made no sense.
Also, is there a quick way to tell which multiplications will produce a coset that I've already obtained?
Link included is my attempt at using WolframAlpha:
https://www.wolframalpha.com/input/?i=permutation+%281+3%29%281+4+2+3%29
Best Answer
You made a mistake while calculating the product of (1 3)(1 4 2 3)
1$\rightarrow$4
4$\rightarrow$2
2$\rightarrow$3$\rightarrow$1
3$\rightarrow$1$\rightarrow$3
The answer should be (1 4 2)
If you want a faster way to calculate cosets, i'll have to direct you to this comprehensive answer that had helped me too!