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

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.

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.
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 $$n+1$$.