How to express any element of $A_n$ I know that $A_n$ is generated by $3-$ cycles; But if I take any element from $A_n$ how can I write it ?
Is that any element of $A_n$ can be expressed as a product of $3-$ cycles that means is it true that any element of $A_n$ is of the form
$(a_1 a_2 a_3)(a_4,a_5,a_6)$ or $(a_1a_2a_3)(a_1a_4a_5)$ or $(a_1a_2a_3)(a_1a_2a_4)$ or$(a_1a_2a_3)(a_2a_1a_4)$ or $(a_1a_2a_3)$
 A: The construction is given by the following proof (outline):
We know that any permutation can be expressed as a product of transpositions and we know how to do this. If $g \in A_n$, then $g$ is an even permutation. As transpositions are odd, this means that $g$ can be expressed as a product even number of transpositions.
Now we look at all the possible products of two transpositions (comes down to three cases) and show how to express each pair as a product of (in fact at most three) $3$-cycles.
Edit: We are clearly not unanimous on what the question actually is... Here's what I answered to. If you are given an element of $A_n$, say $(1 \ 2)(3 \ 4)$ or $(1 \ 2 \ 3 \ 4 \ 5) \in A_5$, then the above construction tells you how to express the element as the product of some finite number of $3$-cycles.
A: No, it is not true. The group $A_n$ is generated by $3$-cycles, but this does not say that every element is the product of only one or two $3$-cycles. For example, consider the element $(123)(456)(789)$ in $A_9$. For a detailed discussion see also How to show that the 3-cycles $(2n-1,2n,2n+1)$ generate the alternating group $A_{2n+1}$.
