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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.

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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 $S$.

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$\langle S \rangle$ is not $S$ itself, it is the group that is generated by the elements of $S$. Every element in $\langle S \rangle$ is a product of elements of $S$. (by definition should be product of elements of $S$ and their inverses but here the inverse of an element of $S$ here is also in $S$ so it doesn't matter). Now, $\tau_1\tau_2, \tau_3\tau_4,\dots,\tau_{k-1}\tau_k$ are all elements in $S$ and $\alpha$ is their product. Hence $\alpha\in\langle S \rangle$.

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The notation $\langle S \rangle$ denotes the group generated by the set $S$ which is a subset of some group $G$. This means the proof has achieved it's objective: it's shown that $\alpha$ is a product of elements in $S$, ie elements that are products of two transpositions.

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By the way, if you use a proof of the simplicity of $A_n$ that does not depend on $A_n$ being generated by $3$-cycles (and the easiest proof, for example, does not depend on it...get simplicity of $A_5$ from the sizes of its conjugacy classes and for larger $n$ use induction and easy standard results on multiply-transitive permutation groups), then you have a simple way to show that $A_n$ is generated by $3$-cycles, for $n \ge 5$. Since the $3$-cycles form a conjugacy class (you don't even need this....just that they are a union of conjugacy classes), they generate a normal subgroup. Since $A_n$ is simple, that subgroup must be $A_n$.

For $n=3$ or $4$, the proof is also easy: The number of $3$-cycles exceeds $|A_n|/2$ in those cases.

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