I was given the following problem:
Consider a pack of $2 n$ cards, numbered from 0 to $2 n − 1$. An outer perfect shuffle is a shuffle of the cards, in which one first splits the pack in two halves of equal sizes and then interleaves the cards of the two halves in such a way that the top and bottom card remain in the top and bottom position. Show that the order of the outer shuffle is the multiplicative order of 2 modulo $2n-1$.
I find it difficult to describe outer shuffling using mathematical language, and I have no knowledge about multiplicative order. How should I proceed?
So we know that the card numbered 0 ends up at position 0. Then underneath that you have the top card from the second half of the deck – that's card $n$ – then you have the second card from the first half, which is card $1$.
You can start to build up a function $f$ such that $f(0) = 0$, $f(1) = n$, $f(2) = 1$, and so on. $f(n)$ is the label of the card in location $n$ in the shuffled deck.
This $f$ is a permutation – to find its inverse, just look at where in the deck the card with label $n$ ended up. The order of the outer shuffle is the minimal $n$ such that $f$ applied $n$ times acts as the identity function.
You can explicitly compute $f$ easily – use a literal pack of actual cards if necessary – and you'll probably only have to compute a handful of values before you start to notice a pattern. Once you have that pattern, try to come up with a simple algorithm for computing $f$ without having to do a whole shuffle, then see if you can come up with a simple description of the function you get by e.g. doing $f$ twice, or three times.
Note also that the order of $f$ is the same as the order of $f^{-1}$, so you can study that as well if it seems more convenient.