Parity of product of all permutations in $S_n$

Question

Suppose $\alpha=$ the product of all permutations in $S_n$ for some $n$. For what $n$ is $\alpha \in A_n$, where $A_n$ denotes the set of all even permutations?

Looking at $S_3$, I've determined that $n=3$ satisfies the condition, since $(1 2 3)(12)(13)(23)(321)$ = $(213)$ which is a product of $3-1=2$ transpositions.

I'm thinking that this question can be answered by examining $k$-cycles for $k>2$ since each permutation and its inverse are the product of the same transpositions when decomposed, and a transposition is the inverse of itself.

Answer 1

1. Because of the fact that only odd permutation in product changes "odd/even" of $\alpha$ then we have to count the number of odd permutations.

2. There are the same amount of odd and even permutations in $S_n$.

3. Because of the fact that $|S_n| = n!$ and $\frac{n!}{2}$ is even for $n \geq 4$ then in $S_n, n\geq 4$ we get that $\alpha$ is even. Otherwise it is odd.

Answer 2

Notice that the product $\alpha$ is not well-defined (since $S_n$ is not commutative when $n>2$), but it's signature is since it takes values in the abelian group $\{\pm 1\}$.

Let $n>2$. There are $n!/2$ permutations with signature $1$ and $n!/2$ permutations with signature $-1$. The product of these signatures is $(-1)^{n!/2}$. If $n=3$, then $n!/2=3$ is odd, but otherwise $n!/2=3 \cdot 4 \cdot \dotsc \cdot n$ is even, in fact it has $4$ as a factor.