The following proof (up until the point I got stuck) was given in my class notes
Proof: Let $S$ denote the set of elements in $S_n$ which are the product of two transpositions. Then $\langle S \rangle \subseteq A_n$. Now if $\alpha \in A_n$ then $\alpha = \tau_1 \tau_2 \dots \tau_k$ where the $\tau_i$'s are transpositions and $k$ is even. Now $\alpha = (\tau_1\tau_2)(\tau_3\tau_4) \dots (\tau_{k-1}\tau_k)$ implies that $\alpha \in \langle S \rangle$. Hence $A_n = \langle S \rangle$.
The last line is the part of this proof that I don't understand. For $\alpha$ to be an element in $\langle S \rangle$ we need to show that $\alpha$ is the product of two transpositions, however in the above I don't see how $\alpha$ is the product of two transpositions at all. We'd need to show that $\alpha = f g$ where $f, g \in S_n$ are both $2$-cycles and I don't see how the last line implies that.
Best Answer
Careful here: you need to show that $\alpha\in\left<S\right>$, not that $\alpha\in S$.
$\left<S\right>$ is the subgroup generated by $S$, so it suffices to show that $\alpha$ is a product of elements of $S$. Thus, $$ \alpha = \tau_1\tau_2\cdots\tau_k = (\tau_1\tau_2)(\tau_3\tau_4)\cdots(\tau_{k-1}\tau_i) $$ is in $\left<S\right>$ because each of $\tau_1\tau_2$ and $\tau_3\tau_4$, etc. are elements of $S$.