Every element in An is product of n-cycles

Question

I just did the following: if $n$ is greater or equal to $3$ than every element in An is product of $3-$cycles. So I suppose this should be a general case. But I did the first problem by dividing it in two special cases and I fail to see how I could use this to prove general statement for $n$ that is : every element in An is product of n-cycles.

Answer 1

We know that every permutation is a product of $2$-cycles, and that an even permutation is a product of an even numbers of permutations. If we can show that a product of two $2$-cycles is equal to a product of $3$-cycles, then the claim is proven.

Let $(a\; b)(c\; d)$ be a product of two $2$-cycles. If the two cycles are equal, then this is trivial. Else, we have $a\neq b,$ $c\neq b$ and $\{a,b\} \neq \{c,d\}$. Without loss of generality, we can take $a\notin \{c,d\}$ and $c\notin\{a,b\}$. Then, you can check that : $$(a\;b)(c\;d) = (a\;c\;b)(c\;d\;a)$$

Remark : this property is true (though trivial) for $n \leq 2$ since we have $\mathfrak A_1 \simeq \mathfrak A_2 \simeq \{1\}$ and the identity is the empty product (of $3$-cycles).

Edit

If $n\geq 4$, every $3$-cycle in $\mathfrak S_n$ is a product of $n$-cycles. Let $\{a_1,\ldots,a_n\} = \{1,\ldots,n\}$ be any ordering. Then, you have : $$(a_1\;a_2\;a_3) = (a_2 \; a_1 \; a_3 \; a_4\; a_5\;\ldots\;a_n)(a_n\;a_{n-1}\;\ldots \;a_2\;a_1)$$ so the claim is a consequence of the property with $3$-cycles.