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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.

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  • $\begingroup$ The product you chose as example from $S_3$ evaluates to $(23)$ (which then is consistent with Martin's answer). $\endgroup$ – vuur Oct 20 '14 at 17:55
  • $\begingroup$ Thanks for clarifying that, but do you mind explaining how that product equals $S_3$? From left to right I see that the product evaluates to $(23)$, but from right to left I get $(213)$. $\endgroup$ – Seh-kai Liao Oct 20 '14 at 18:04
  • $\begingroup$ I'm not used to calculating from left to right, but I think it goes like this: [123]->[312]->[213]->[231]->[132]->[213] = (12) (where the numbers in square brackets is the set acted on and each arrow corresponds to one of the five cycles read from the right acting on the set) Note that after the last action the result must be converted to a cycle. You should ask: "Which cycle alone produces the same result as the five cycles"? $\endgroup$ – vuur Oct 20 '14 at 18:19
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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.

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  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.

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