I have a problem with permutations (involutions): If I have the set $$S=\{1,2,\dots,26\}$$ how can I find a function $f$ that assigns any $n$ from a subset of the natural numbers (lets call it $N$) a unique involution $\sigma$ of $S$? Also how can I find a function $g$ that does the inverse, so that $f(g(\sigma))=\sigma$ and $g(f(n))=n$?
Im searching for a method that allows me to compute the involution from $n\in N$ as well as compute $n$ from the involution.
I'm going to take permutations of $S'=\{0,1,\dotsc,25\}$ because it's easier on the notation. Let the set of involutions on this set be $J$.
Take the integers in base $26$. Let $\sigma$ be an involution on your set. Define an integer in the following fashion: take $$ g(\sigma)=\sum_{n=0}^{25} \sigma(n) 26^n $$ This gives you a 26-digit number in base $26$, which uniquely corresponds to the permutation $\sigma$, because different permutations have $\sigma(n)$ different for some $n$, and hence one of the digits in their images under $g$ is different. It's also clear how to extract a permutation from this representation: $\sigma(n)$ is the coefficient of $26^n$.
Okay, that's the easy bit. What is the set we're mapping to? Well, it's the subset of $\{0,1,2,\dotsc,26^{26}-1\}$ where all the digits are different in base $26$, and the $n$th digit is $m$ if and only if the $m$th digit is $n$ (i.e. if we have $n 26^m$ in the above expansion, we also have $m 26^n$) (It's easy to see why since we need $\sigma^2=\text{id.}$ Yeah, it's not a very nice set to describe.