TITLE: Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$
QUESTION [5 upvotes]: Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$

My progress:
Consider any $S={(x,y),~~1\le x\le p,~~1\le y\le 2}.$ Note that $|S|=2p.$
Now consider $A\subseteq S$ such that $|A|=p.$
So there are a total of ${2p}\choose{p}$ possible value of $A.$
Then let $A=\{(a_1,b_1),......,(a_p,b_p)\}.$
Define $$f(A)=\{(a_2,b_2),.............(a_p,b_p),(a_1,b_1)\}$$
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$$f^{p-1}(A)=\{(a_p,b_p),\dots ,(a_{p-1},b_{p-1})\}$$
Now, define a new binary relation $R$ such $f^{m}(A)Rf^l(A)$ when $$b_{m+1}=b_{l+1},~~b_{m+2}=b_{l+2},~~\dots$$
Note that this is possible only when $b_i=c$ a constant where $c=1$ or $2.$
For the rest ${{2p}\choose{p}}-2$ sets, the orbit size is $p.$
Hence we get $p\mid{{2p}\choose{p}}-2.$
I want to go with the same essence, but I don't know how to get $p^2.$ Any hints?

REPLY [3 votes]: I'm assuming $p$ is prime.
To paraphrase your work so far, you have a $2\times n$ array of points and you're acting by rotating the $n$ columns cyclically. Instead, rotate the $n$ points in each row independently.
That is, let $f$ map $(x,1)$ to $(x+1\text{ mod }p,1)$ while fixing the points $(x,2)$, let $g$ map $(x,2)$ to $(x+1\text{ mod }p,2)$ while fixing the points $(x,1)$, and consider the action of the group $\langle f,g\rangle$ on $\{A\subset S : |A|=p\}$. Then for any $A$ other than $\{(x,1):1\le x\le p\}$ and $\{(x,2):1\le x\le p\}$, the orbit of $A$ has size $p^2$.
(This is closely related to robjohn's answer, but perhaps sticks closer to the spirit of actions and orbits.)