Prove that ${2p\choose p}\equiv 2\pmod {p^2}.$ 
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?
 A: Vandermonde's Identity says that
$$
\binom{2p}{p}=\sum_{k=0}^p\binom{p}{k}\binom{p}{p-k}\tag1
$$
For $k\ne0\pmod{p}$, we have $\left.p\,\middle|\binom{p}{k}\right.$, thus $\left.p^2\middle|\binom{p}{k}\binom{p}{p-k}\right.$ for all $k$ except $0$ and $p$.
$\binom{p}{k}\binom{p}{p-k}=1$ for $k=0$ and $k=p$.
A: $$\binom{2p}p=\frac{2p(2p-1)\cdots(p+1)}{p(p-1)\cdots 1}=2\frac{(p+1)(p+2)\cdots(2p-1)}{1\cdot 2\cdots (p-1)}$$
Now, note that $(p+k)(p-k)\equiv -k^2\pmod{p^2}$ for $1\le k\le p-1$; also $\gcd(k^2,p^2)=1$. Basically, for any $n$ such that $p\not\mid n$, we have $\gcd(n,p^2)=1$ and we can divide num/denom by it.
Hence,
$$\prod_1^{p-1}\frac{p+k}{p-k}\equiv\prod_1^{p-1}\frac{(p+k)(p-k)}{(p-k)^2}\equiv\prod_1^{p-1}\frac{-k^2}{(p-k)^2}\equiv (-1)^{p-1}\pmod{p^2}$$
whence we have $\binom{2p}p\equiv 2(-1)^{p-1}\pmod{p^2}$
For prime $p\gt 2$, $(-1)^{p-1}\equiv 1$ and the claim follows. For $p=2$, we have $-2\equiv 2\pmod{2^2}$, so the claim holds there as well.
Thus, $\binom{2p}p\equiv 2\pmod{p^2}$ for all primes $p$
A: 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.)
