# Permutation and symmetric Group

For all $i,j \in \{1,2,3,4\}$ such that $i+j=5$,

let $G$ be the set of permutations $\sigma \in S_4$ satisfying $\sigma (i) + \sigma(j)=5$.

a.) List all the elements of G and write a multiplication table for G. (Convention: The product of two permutations is given by composition)

b.) Let H be the subgroup of G consisting of the identity element and the permutation (2,3). Determine, with the help of the multiplication table, the left H-cosets and the right H-cosets. Is every left H-coset a right H-coset?"

First I'm not sure if I understand the task correctly. I did write down the tuple satisfying $$i+j=5$$ which are: (1,4),(2,3),(4,1),(3,2). Do I now have to insert them into the function $$\sigma (i) + \sigma(j)=5$$ and list a possible permutation like showing in the following two examples: $$\sigma (1) + \sigma(4)=5$$ with permutation (12)(43) or $$\sigma (1) + \sigma(4)=5$$ with permutation (14)(23)? If yes, then I would find these example easily: (12)(43),(43)(12),(24)(31),(31)(24). But still i'm confused how I should handle permutations of solely 2 digits like (14)? I mean if I have (14)(3)(2) I could still pick 2 and 3 to satisfy the condition. So I actually do not no, when the permutation does not satisfy the condition..

And how do I write a multiplicationtable if it's not commutative? Do I start with the column and then comes the row? Example ("row3 ° col5" or "col5 ° row3")

Thanks for help

• It really doesn't matter how you compose them (last question), as long as you are consistent. If you first create a table composed with row $\circ$ col, and then do the same composed with col $\circ$ row, the tables will be equivalent, up to isomorphism. – Namaste Oct 26 '14 at 15:42
• Thank you, any chance if you could check whether I understood the task correctly? – Matriz Oct 26 '14 at 16:19
• Disjoint permutations (permutations with disjoint set of moved points) always commutate. So for example $(1 2)(4 3) = (4 3)(1 2)$. – user Oct 29 '14 at 15:47
• $(1 4)$ is in $G$. Just check for all $i$ and $j$. – user Oct 29 '14 at 15:51