Alternative proof that the 3-cycles generate the alternating group $A_n$

Question

A lot of proofs I have seen involve writing $p \in A_n$ as a product of an even number of transpositions. I have a different proof that involves disjointedness like in my question earlier today Prove $(\rho_1 \rho_2 \cdots \rho_r)^u = e \implies e=\rho_1^u=\rho_2^u=\cdots=\rho_r^u$.

Let $p \in S_n$.

• If $p$ is a cycle, then $sign(p)=1 \iff p \in A_n \iff length(p)$ is odd, which can be seen from computing $\det(P)$, where $P$ is the associated permutation matrix to $p$. Write $p = (a_1 ... a_{2m+1}) = (a_1 a_2 a_3)(a_3 ... a_{2m+1})$ where $(a_3 ... a_{2m+1})$ has odd length because $(2m+1)-(3)+1=2m-1$, which is odd. Eventually, we can write every cycle of odd length as a product of 3-cycles.

• If $p$ is not a cycle, then let its disjoint cycle decomposition be $p=\rho_1 \rho_2 ... \rho_r$.

• If they are all cycles of odd length, then we are done.

• If one has a cycle of even length, call it $\rho_{s_1}$. If $\rho_{s_1}$ is a 2-cycle, then by disjointedness, no other $\rho_{s_i}$ returns the switched indices, even though there is some other cycle of even length, call it $\rho_{s_1}$ (because $\det(\prod_{i=1}^m P_i) = \prod_{i=1}^{m}\det(P_i) = \prod_{i=1}^{m}(1) = 1$ implies that negative $\det(P_i)$'s come in pairs) and therefore, $p$ is somehow odd, a contradiction.

• I also haven't worked out what to do if $\rho_{s_1}$ is an odd cycle of length 4,6,8, etc.

Any suggestions on how to continue?

Note: No using $(a b)(c d)=(a d c)(a b c)$ and $(a b)(b d) = (a b d)$, though I'm starting now to see why all the proofs I've found involve this.

Answer 1

I would prove it somewhat differently.

If $A_n$ are permutations acting on a set $\{1,2,...,n\}$, then you can consider $A_{n-1}$ to be a subgroup of $A_n$. It is all the permutations that don't move item $n$.

If you have any element $p\in A_n$, then either it does not move item $n$ and we have $p\in A_{n-1}$, or it does move item $n$ and we can find a 3-cycle $c$ to undo that move so that $cp\in A_{n-1}$. Repeat this procedure until you get down to $A_3$ and you will have expressed $p$ as a product of 3-cycles.

The above is a descent approach, top-down. You can also use a bottom-up approach to prove it more formally via induction.

We want to prove that all $n$, $A_n$ is generated by the $3$-cycles of the form $(12k)$, where $3\le k \le n$. Show it is true for $n=3$, and then show that if it is true for any particular $n$, then it follows that it also is true for $n+1$.