I just have small question about the proof. In order to show that $A_{n}\leq S_{n}$ one has to (among other things) show that the identity element $\varepsilon$ in $S_{n}$ is also the identity element in $A_{n}$. Now, when I'm reading of the notes from my professor he just writes
Identity: Clear, since $\varepsilon =\bigl( 1 2 \bigr)\bigl( 2 1 \bigr)$.
I dont understand what he means when he writes this. I would instead argue that $\varepsilon \in S_{n}$ is always an even permutation since it can be written as $0$ transpositions and $0$ is even. Thus $\varepsilon\in A_{n}$. Is this a valid argument?
Both answers are fine. I think that your justification is the better one, but you have to admit that, indeed, $\varepsilon=(1\ \ 2)(2\ \ 1)$ and that this equality also proves that $\varepsilon\in A_n$. Also, some beginners may find it hard to understand the idea that $\varepsilon$ is the composition of $0$ transpositions.