Question on proof: cosets of the alternating group $A_n$

Question

The theorem:

If $n \ge 2$, then $|A_n| = \frac{n!}2$.

Proof:

We will proof that $S_n\backslash A_n$ is the only non-trivial coset of $A_n$ (so aside from $A_n$ itself).

Consider an arbitrary, but fixed transposition $\phi$: $A_n$ does not contain transposition, thus $\phi \in S_n\backslash A_n$. Choose another element $\theta \in S_n \backslash A_n$. We know that $\det(M_{\phi})=\det(M_{\theta})=-1$, thus $\det(M_{\phi^{-1}}M_{\theta})=1$, meaning that $\phi^{-1}\theta \in A_n$. Eventually we get $\theta \in \phi A_n$.

Now we've proved that $|S_n\backslash A_n|=|A_n|$; both cardinalities are therefore half of $|S_n|=n!$.

My question:

I don't understand the last sentence. How does it follow from $\theta \in \phi A_n$. Also how does this imply that $S_n \backslash A_n$ is the only non-trivial coset of $A_n$? We could have taken another transposition instead of $\phi$, right?

Thanks in advance.

Answer 1

What the proof actually shows is that $S_n\setminus A_n\subseteq \phi A_n$.

But $S_n = A_n \stackrel{\cdot}{\cup} (S_n\setminus A_n)$ and $A_n\cap \phi A_n=\emptyset$, and so $S_n\setminus A_n=\phi A_n$. The sets $A_n$ and $\phi A_n$ have the same cardinality by the bijection $\tau\mapsto \phi\tau$ and so $A_n$ has cardinality $n!/2$.

Answer 2

The proof shows that for any element $\theta\in S_n\setminus A_n$ we have $\theta\in\phi A_n$; left-multiplication by $\phi$ shows that $$\phi^{-1}\theta \in A_n\quad\implies\quad\theta\in\phi A_n.$$ It follows that $S_n\backslash A_n$ is a subset of $\phi A_n$, where $\phi A_n$ is a coset of $A_n$. This implies that $$|A_n|=|\phi A_n|\geq|S_n\backslash A_n|=|S_n|-|A_n|,$$ and a bit of rearranging shows thatt $|A_n|\geq\tfrac{|S_n|}{2}$. Of course $|A_n|<|S_n|$ and so it follows that $$|A_n|=\tfrac{|S_n|}{2}=\frac{n!}{2}.$$ You are correct in saying that we could have chosen any other transposition $\phi'\in A_n$ to reach the same conclusion. In fact, this shows that $\phi A_n=\phi'A_n$ for any pair of transpositions $\phi,\phi'\in S_n$. And this makes sense, because their 'difference' $\phi^{-1}\phi'$ is an even permutation, meaning that $\phi^{-1}\phi'\in A_n$.