Proof that the alternating group is generated by three-cycles

Proof

We show two inclusions. Let $n \geq 3$, let $T$ be the set of $3$-cycles in $S_n$, and let $H = \langle T \rangle$ be the subgroup generated by $T$. Products are composites of maps, and the right factor acts first.

1. By the sign of a cycle, each $3$-cycle has sign $(-1)^{2} = 1$. So $T \subseteq A_n$, by the definition of the alternating group.
2. The set $A_n$ is the kernel of the sign homomorphism, so it is a subgroup of $S_n$. Since $H$ is the smallest subgroup that contains $T$, step 1 gives $H \subseteq A_n$.
3. For the other inclusion, we first show that each product of two transpositions lies in $H$. Let $\alpha$ and $\beta$ be transpositions.
4. If $\alpha = \beta$, then $\alpha \beta = e$, and $e \in H$.
5. If $\alpha$ and $\beta$ move exactly one common point $a$, write $\alpha = (a\ b)$ and $\beta = (a\ c)$ with $a, b, c$ distinct. Then $\alpha\beta$ sends $a \mapsto c$, $c \mapsto b$ and $b \mapsto a$, and fixes the other points.
6. So in cycle notation $\alpha \beta = (a\ c\ b)$, which lies in $T \subseteq H$.
7. If $\alpha$ and $\beta$ move no common point, write $\alpha = (a\ b)$ and $\beta = (c\ d)$ with $a, b, c, d$ distinct. Let $\gamma = (a\ c\ b)(a\ c\ d)$.
8. Compute $\gamma$ point by point: $a \mapsto c \mapsto b$, $b \mapsto b \mapsto a$, $c \mapsto d \mapsto d$, $d \mapsto a \mapsto c$. Other points are fixed.
9. So $\gamma = (a\ b)(c\ d) = \alpha \beta$. It is a product of two elements of $T$, so it lies in $H$.
10. Steps 4 to 9 cover all cases, because two transpositions move two points each and share zero, one or two points. Sharing two points means $\alpha = \beta$.
11. Now let $\sigma \in A_n$. By the decomposition into transpositions, $\sigma = \tau_1 \tau_2 \cdots \tau_m$ with transpositions $\tau_i$.
12. By the definition of the sign, $(-1)^m = \operatorname{sgn}(\sigma) = 1$, so $m$ is even. Write $m = 2r$.
13. Then $\sigma = (\tau_1 \tau_2)(\tau_3 \tau_4) \cdots (\tau_{2r-1} \tau_{2r})$. Each bracket lies in $H$ by steps 4 to 10, and $H$ is closed under products. So $\sigma \in H$.
14. Thus $A_n \subseteq H$. With step 2, $A_n = H$, so $A_n$ is generated by the $3$-cycles.