Number of odd and even permutation.

Question

I needed to find number of odd and even permutations in a symmetric group $S_n$ (having $n$ elements).

What we do is select a arbitrary fixed odd permutation $h \epsilon S_n $. We know that $hS_n=${$hg:g\epsilon S_n$} = $ S_n$.

Let's say there are $x$ odd permutations and $y$ even permutations in $S_n$.

Number of odd permutation in $hS_n$ is $y$ (formed $h \cdot \#$ (even permutation $\in S_n$)) and even permutation=$x$.

As $S_n$ and $hS_n$ are same sets therefore $x=y$. Therefore even = odd = $\frac{n!}{2}$. So , according to my proof if we can find even one odd permutation then number of even and odd permutation are same in any group. A subset of $S_n$ can only be a subgroup of $S_n$ if the subset either contains equal odd and even permutations or only even permutation.

Is my result correct, or did I make any mistake?

Answer 1

Okay, you made some mistakes but unfortunately, I can't edit your question. But I assume you mean that the map :$\sigma \mapsto h* \sigma$ is a bijective map from $S_n$ to $S_n$ that maps even permutations to odd and odd permutations to even. in that case, you are correct.

also try to think of a homomorphism from $S_n$ to $S_2$ ,what does the kernel tell you.

Answer 2

By way of enrichment I would like to show how to solve this using analytic combinatorics. We have that the sign of a permutation $\pi$ is given by

$$\sigma(\pi) = \prod_{c\in \pi} (-1)^{|c|-1}$$

where $c$ iterates over the cycles of the permutation and $|c|$ is the length of the cycle. Therefore we use the following combinatorial class of permutations with the length of cycles minus one marked:

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}( \textsc{CYC}_{=1}(\mathcal{Z}) + \mathcal{U}\times \textsc{CYC}_{=2}(\mathcal{Z}) + \mathcal{U^2}\times \textsc{CYC}_{=3}(\mathcal{Z}) + \mathcal{U^3}\times \textsc{CYC}_{=4}(\mathcal{Z}) + \cdots).$$

This gives the EGF

$$G(z, u) = \exp\left(z+u\frac{z^2}{2} + u^2\frac{z^3}{3}+ u^3 \frac{z^4}{4}\cdots\right) \\ = \exp\left(\frac{1}{u}\log\frac{1}{1-uz}\right).$$

It follows that the EGF of even permutations is given by

$$H(z) = \frac{1}{2} (G(z,1)+G(z,-1)) \\ = \frac{1}{2} \left(\exp\log\frac{1}{1-z} + \exp\left(-\log\frac{1}{1+z}\right)\right) \\ = \frac{1}{2} \left(\frac{1}{1-z} + 1+z\right).$$

Therefore we have for $n\ge 2$

$$n! [z^n] H(z) = \frac{1}{2} n!$$

and as boundary cases the value one for $n=0$ and $n=1$ (these two contain zero transpositions).