Cosets of alternating group in symmetric group

Question

Let $G=S_n$ and $N=A_n$. There are exactly two left cosets of $A_n$ in $S_n$: $1\cdot A_n$ and $\left(1,2\right) \cdot A_n$, the latter consisting precisely of all odd permutations. Thus, the quotient $S_n/A_n$ is cyclic of order 2

1. I think the left cosets are found using $ \left(\left(1,2\right)A_n\right)^2=\left(1,2\right)^2\cdot A_n=1\cdot A_n$ and the other one $(1,2)\cdot A_n$ is basically $S_n-A_n$, i.e. $xA_n$ for all $x \notin A_n$ but not sure how to justify these.

2. Not sure how the quotient is cyclic.

3. Is the quotient of order 2 because the 2 left cosets have equal orders?

Apologies for the messy format... The proof seems trivial but I am really not sure how to prove this. Any help will be very much appreciated!

Answer 1

1. The set of left cosets is: $$\mathcal{L}=\{aA_n | a\in S_n\}.$$ Now it is easy to see that $1 \cdot A_n=A_n$ and $(1,2) A_n$ will give you a full list of the members of $\mathcal{L}$. In fact, you always have $H$ in $G/H$ so this implies that $1\cdot A_n \in \mathcal{L}$. Now, if you multiply $A_n$ by $(1,2)$ (or any odd permutation), then you are considering the elements of the form $ (1,2) \cdot \text{(even permutation)}$. The even permutation will give you an even number of transpositions and you have one more which is (1,2), so this takes you to odd permutations. Thus, $(1,2) A_n \subset S_n - A_n = B_n$.

Indeed, $|(1,2)A_n|=|A_n|=|B_n|$. Hence, $(1,2)A_n = B_n$.

By a similar treatment you can show that $(\text{even permutation}) \cdot A_n = A_n$ and $(\text{odd permutation}) \cdot A_n = B_n$.

Therefore, $1 \cdot A_n$ and $(1,2) \cdot B_n$ gives you the required full list.

2. Any group of prime order is cyclic.

3. The order of the quotient is $|S_n/A_n|=|S_n|/|A_n| = \dfrac{n!}{n!/2}=2.$