Let $m\leq n−2$. Given any $m$ distinct elements $i_1,...,i_m$ from $\{1,...,n\}$, show that there exists a $σ ∈ A_n$ such that $$\sigma(j) = i_j,\ \ \ \forall j\in \{1,...,m\}$$
What I could do
I have done almost nothing. I think we have to do induction here. It is easy to show for $n=3$ case, but the inductive step is where I am stuck. Please help.
Assume first $m=n-2$. Take any permutation $\sigma\in S_n$ which satisfies $\sigma(j)=i_j$. Then the pair $x,y$ of numbers among $\{1,2,\dots,n\}$ that does not belong to the initial choice of $n-2$ numbers is mapped to itself. That is, either $\sigma(x)=y, \sigma(y)=x$ or else $\sigma(x)=x,\sigma(y)=y$. Now, if $\sigma$ is already even, stop; else, if $\sigma$ is odd, consider the permutation $\tilde{\sigma}$ that agrees with $\sigma$ on $1\leq j\leq m$ but does the opposite on the pair $\{x,y\}$. This $\tilde{\sigma}$ would then be even.
A simple modification works for $m<n-2$.