Vtgyjfy

We have a stack of $n$ books piled on each other, and labeled by $1, 2, ..., n$. In each round we make $n$ moves in the following manner: In the $i$-th move of each turn, we turn over the $i$ books at the top, as a single book. After each round we start a new round similar to the previous one. Show that after some moves, we will reach the initial arrangement.

Say $n=4$ and initial arrangement of books $(a,b,c,d)$. First we act on it with identical transformation $pi_1=id$ which leaves everything as it was. Then we act on it with
$$pi_2 = left(
begin{array}\
1 & 2 & 3 & 4 \
2 & 1 & 3 & 4
end{array}right)$$
and we get $(b,a,c,d)$, then we act on this one with $$pi_3 = left(
begin{array}\
1 & 2 & 3 & 4 \
3 & 2 & 1 & 4
end{array}right)$$ and we get $(c,a,b,d)$ and then $$pi_4 = left(
begin{array}\
1 & 2 & 3 & 4 \
4 & 3 & 2 & 1
end{array}right)$$ and we get $(d,b,a,c)$ and then we repeat acting with $$pi_1,pi_2,pi_3,pi_4,pi_1,pi_2,...$$

Now what do we get if we repeat enough time $$sigma = pi_4circ pi_3circ pi_2circ pi_1$$ on starting $(a,b,c,d)$? If we repeat this $sigma $ exactly $24$ times (which is the order of symmetric group $S_4$) we shoud get initialy arrangment. Clearly this can be easly generalized for arbitrary $n$.

Is this correct?

Edit: As suggested in comment by Lord Shark the Unknown, It shoud be considered also the first and last front of a book. So I should observe 8-couple $(a1,a2,b1,b2,c1,c2,d1,d2)$, where $x1$ is first front and $x2$ last one, instead of 4-couple $(a,b,c,d)$ and act on it with $S_8$?

Let $S$ be the set of $2n$ book covers. Each round is a permutation of $S$. Every permutation $pi$ of a finite set has a number $k$ such that $pi^k=text{id}$. One can choose $k$ to be the lcm of the cycle lengths of $pi$.

For example, when $n=2$, a single round looks like this, where capital letters are the top cover and lower case are the bottom cover.

A    a    b

a    A    B

B    B    A

b    b    a

In cycle notation, this looks like $(A;; B;; a;; b)$. This a cycle of order four, so four rounds suffice to return the books to their original order.

For clear reference, here is a complete cycle of moves. Negative number represents book face down.

 1  2  3  4 

-1  2  3  4 

-2  1  3  4 

-3 -1  2  4 

-4 -2  1  3 

 4 -2  1  3 

 2 -4  1  3 

-1  4 -2  3 

-3  2 -4  1 

 3  2 -4  1 

-2 -3 -4  1 

 4  3  2  1 

-1 -2 -3 -4 

 1 -2 -3 -4 

 2 -1 -3 -4 

 3  1 -2 -4 

 4  2 -1 -3 

-4  2 -1 -3 

-2  4 -1 -3 

 1 -4  2 -3 

 3 -2  4 -1 

-3 -2  4 -1 

 2  3  4 -1 

-4 -3 -2 -1 

 1  2  3  4 << return to initial state