What are the left and right cosets of $A_n$ in $S_n$?

I know that $A_n$ is the set of all even permutations and $S_n$ is the set of all permutations.

I was looking online and found that since $A_n$ is normal, the other coset is $S_n - A_n$(since cosets partition a group).

For $A_4$ in $S_4$, it's pretty visual because you can clearly see $e$ and $(1,2)$ are the left cosets and $e$ is the right coset.

Is there any way to visually see it for $A_n$ in $S_n$ by listing?

Thanks in advance.

  • 6
    $\begingroup$ Welcome to Mathematics Stack Exchange. $e$ and $(1,2)$ are representatives of cosets, and since $A_n$ is a normal subgroup, the right and left cosets are the same $\endgroup$ Commented Nov 5, 2019 at 1:49
  • $\begingroup$ What do you mean by "visually see"? $\endgroup$
    – xxxxxxxxx
    Commented Nov 5, 2019 at 16:30
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    – Shaun
    Commented Nov 5, 2019 at 16:41

1 Answer 1


Since $A_n$ is normal in $S_n$, we have for all $\sigma\in S_n$ that $\sigma A_n\sigma^{-1}=A_n$; hence $\sigma A_n=A_n\sigma$; that is, the left and right cosets are the same.

Also, here $e$ and $(1,2)$ are not cosets; they are representatives of cosets. The cosets look like this:

$$\sigma A_n=\{\sigma\tau\mid \tau\in A_n\}$$

for each $\sigma\in S_n$ up to the representative.

To see $A_n$ in $S_n$ visually, try reading "Visual Group Theory," by Carter.


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