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Here's a question I'm working on:

Show that if $\rho$ and $\sigma$ are any two permutations, then

$(a)$ $\sigma\rho$ and $\rho\sigma$ have the same parity and

$(b)$ $\sigma$ and $\rho\sigma\rho^{-1}$ have the same parities

For the first part of the proof, I don't know how to generalize the result to any permutation, but I do know that, let's say $$\rho = \begin{pmatrix} 1 & 2 & 3& 4\\3 & 1 & 4 & 2 \end{pmatrix}$$ and that $$\sigma = \begin{pmatrix} 1 & 2 & 3 &4\\2 & 3 & 4 &1 \end{pmatrix}$$

So $$\sigma\rho = \begin{pmatrix} 1 & 2 & 3 &4\\4 & 2 & 1 & 3 \end{pmatrix}$$ and $$\rho\sigma = \begin{pmatrix} 1 & 2 & 3 & 4\\1 & 4 & 2 & 3 \end{pmatrix}$$

So their parities are given by $$\sigma\rho = (3-1)+(1-1) = 2$$ and $$\rho\sigma = (1-1)+(3-1) = 2$$

For part $(b)$, it's the same situation, except that we work with the inverse of $\sigma$. I would like to know how to generalize my result for any permutations for $\rho$ and $\sigma$.

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John W. Smith
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  • $\rho$ is of even parity iff it is expressable as a product of an even number of transpositions. $\rho$ is of odd parity iff it is expressable as a product of an odd number of transpositions. Have you seen the proof for why a permutation cannot simultaneously be an even and an odd permutation? Letting $\rho = r_1r_2\dots r_n$ and $\sigma=s_1s_2\dots s_m$ be representations in terms of products of transpositions, before cancelling, how many transpositions are in the product $\rho \sigma$? In $\sigma \rho$? – JMoravitz Apr 04 '17 at 20:46
  • You can represent a permutation with a permutation matrix. The determinant of such a matrix is $1$ for even permutations, and $-1$ for odd permutations. Using $det(AB) = det(A)det(B)$ makes short work of the problem. – Kaynex Apr 04 '17 at 20:47
  • To answer your first question, the book does state it as one of the exercises in my textbook, so the author doesn't show a proof of it. – John W. Smith Apr 04 '17 at 20:48
  • For the second part, would $\rho\sigma$ have $n$ transpositions, the same answer goes for $\sigma\rho?$ – John W. Smith Apr 04 '17 at 20:51

1 Answers1

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If you are allowed to use the notion of signature, this is straightforward, because signature is a group homomorphism :

$$\forall (\sigma,\rho)\in\mathfrak S_n^2,\,\epsilon(\sigma\circ\rho)=\epsilon(\sigma)\,\epsilon(\rho)=\epsilon(\rho)\,\epsilon(\sigma)=\epsilon(\rho\circ\sigma)$$

and something analogous for the other equality.

Adren
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  • Unfortunately, I am not familiar with that notation Adren. But it's still good to see how people approach this differently. – John W. Smith Apr 04 '17 at 20:53