Number of odd and even permutation. 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?
 A: 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.
A: 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).
