I want to prove that for $n \geq 3$, the alternating group $A_n$ is generated by the set of all 3-cycles.
Here is my attempt:
Let $\mathcal{S}$ be the set of all 3-cycles in $S_n$, which is a conjugacy class, and let $H=\langle \mathcal{S} \rangle$. For any $\sigma \in S_n$ we have
\begin{equation}
\sigma H \sigma^{-1}=\sigma \langle \mathcal{S} \rangle \sigma^{-1}=\langle \sigma \mathcal{S} \sigma^{-1} \rangle=\langle \mathcal{S} \rangle=H,
\end{equation}
so that $H \trianglelefteq S_n$. By a previous exercise (listing the normal subgroups of $S_n$ for $n \geq 5$), $H \in \{1,A_n,S_n\}$. $H$ is obviously non-trivial, and since all elements of its generating set are even permutations it cannot be $S_n$. We conclude that $\langle \mathcal{S} \rangle=A_n$.
The above argument only works for $n \geq 5$. We are left with handling the cases $n \in \{3,4\}$ separately. Both are pretty easy to verify: If $n=3$, $|A_3|=3$ is cyclic and generated by the 3-cycle $(1 \;2 \;3)$. If $n=4$, we have $8$ 3-cycles in $S_4$. Lagrange's Theorem then forces $\langle \mathcal{S} \rangle=A_4$.
Is my proof correct? If not, please help me fix it.
Thanks!
Best Answer
This proof works.
As @bof noted in a now-deleted comment, there is a more direct approach which works for all $n$, not using the more advanced result about the normal subgroups of $S_n$ for $n>4$. Namely, since $A_n$ is generated by permutations which are the product of two transpositions, it's enough to show that such a permutation is a product of $3$-cycles, e.g., that $(1\ 2)(2\ 3)=(1\ 2\ 3)$ and $(1\ 2)(3\ 4)=(1\ 2\ 3)(2\ 3\ 4)$.