Proving Evenness of m and r in σ ∈ Sn

[namzay300]

Suppose σ ∈ Sn and σ = τ₁*τ₂****τ_m = t₁*t₂****t_r where each τ_i and t_j is a transposition. Then, prove m is even iff r is even.

Note: τ(δ(r₁,r₂,...,r_n) = -δ(r₁,r₂,...,r_n)

Usually like to provide what I have done so far, but I've been racking my brain for awhile and can't come up with much. Any detailed insight would be much appreciated. Thanks!

 

[Deveno]

Suppose σ ∈ Sn and σ = τ₁*τ₂****τ_m = t₁*t₂****t_r where each τ_i and t_j is a transposition. Then, prove m is even iff r is even.

Note: τ(δ(r₁,r₂,...,r_n) = -δ(r₁,r₂,...,r_n)

Usually like to provide what I have done so far, but I've been racking my brain for awhile and can't come up with much. Any detailed insight would be much appreciated. Thanks!

http://mathhelpboards.com/linear-abstract-algebra-14/symmetric-polynomials-involving-discriminant-poly-17823.html

Are you two taking the same class? (Wink)

 

[namzay300]

http://mathhelpboards.com/linear-abstract-algebra-14/symmetric-polynomials-involving-discriminant-poly-17823.html

Are you two taking the same class? (Wink)

That would be quite a coincidence! I was hoping to receive an answer pertaining to my question though :( If anyone out there can help, that would be great! Thank you.

 

[Deveno]

That would be quite a coincidence! I was hoping to receive an answer pertaining to my question though :( If anyone out there can help, that would be great! Thank you.

The link I posted was to another thread that essentially covers the same ground, although it may not seem like it.

The strategy is to show that if $\sigma = (i\ j)$ is a transposition, then the action of $\sigma$ on the Vandermonde polynomial $\delta(x_1,\dots,x_n) = \prod\limits_{k < m} (x_k - x_m)$ given by:
$\sigma(\delta(x_1,\dots,x_n)) = \delta(x_{\sigma(1)},\dots,x_{\sigma(n)})$ is equal to $-\delta(x_1,\dots,x_n)$.

So, by extension, if a permutation $\tau$ can be written as an even number of permutations, it leaves the Vandermonde polynomial unchanged, and if it can be written as an odd number of permutations, it changes the sign of the Vandermonde polynomial.

Now if a permutation could be written as both an even number AND an odd number of permutations, we would have:

$\delta(x_1,\dots,x_n) = -\delta(x_1,\dots,x_n)$, a contradiction.