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?
Best Answer
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.