Number of permutations in $A_n$ which fixes two elements 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!
 A: 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 $|H|=1=(n-2)!$.
A: Denote your group by $H$ and let $X = \{2,4,5,\ldots,n\}$. It is straightforward to check that
$H \to A_X$, $\sigma \mapsto \sigma|_X$ well-defined, bijective and a group homomorphism. So $H \cong A_X \cong A_{\lvert X\rvert} = A_{n-2}$ and therefore
$$
\lvert H\rvert = \lvert A_{n-2}\rvert = \begin{cases} n!/2 & \text{if }n > 3;\\1 & \text{ f }n = 3.\end{cases}
$$
