Number of permutations in $A_n$ which fixes two elements

Question

I want to find the number of permutations in $A_n$, $n\geq 4$ which fix 1 and 3. In $S_n$, the answer is certainly $(n-2)!$. Is it $(n-2)!/2$? I am stuck here. Can we say that $A_{n-2}$ and $\{\sigma\in A_n:\sigma(1)=1,\sigma(3)=3\}$ are isomorphic? Please help!

Answer 1

It is pretty much it, your group is by definition $H:=A_n\cap \mathfrak{S}_{\{2,4,5,...,n\}}$. In particular this is a subgroup of $\mathfrak{S}_{\{2,4,5,...,n\}}$. Now, since $A_n$ is the kernel of a group morphism onto $\{\pm 1\}$ (this the so-called signature), the group $H$ can be seen as the kernel of a group morphism $\phi: \mathfrak{S}_{\{2,4,5,...,n\}}\rightarrow \{\pm 1\}$ (namely the restriction of the signature).

If $\phi$ is not surjective then $H=Ker(\phi)=\mathfrak{S}_{\{2,4,5,...,n\}}$.

If $\phi$ is surjective then $H=Ker(\phi)$ is of index $2$ in $\mathfrak{S}_{\{2,4,5,...,n\}}$. Since in a $\mathfrak{S}_k$ (with $k\geq 2$) there exists a unique subgroup of index $2$ : $A_k$, we see that $H=A_k$.

Now assume $n=3$ then $\mathfrak{S}_{\{2,4,5,...,n\}}$ is the trivial group so $H=A_1$.

If $n>3$ then $\phi((2,4))=-1$ so $\phi$ is surjective and, again, $H=A_{\{2,4,5,...,n\}}$.

If $n>3$ then $|H|=\frac{(n-2)!}{2}$, and if $n=3$ then $|H|=1=(n-2)!$.