Prove that the binomial coefficent $\binom{2p}{p} $ is $\equiv 2\pmod{p}$ using group actions. I have to prove that the binomial coefficent $\binom{2p}{p} $ is $\equiv 2\pmod{p}$ using group actions.
I've tried with an action of $ C_p \times C_p$ upon the set of all numbers between $1$ and $2p$ but I'm blocked here. 
P.s. here $C_p$ is the cyclic group of order p 
 A: We may consider every subset $S$ of $\{1,\ldots,2p\}$ with $p$ elements as a couple $\left(S\cap \{1,p\},S\cap\{p+1,2p\}\right)$, then consider the action (by shifting) of $C_p\times C_p$ on such couples. 
Iff both $S\cap\{1,p\}$ and $S\cap\{p+1,2p\}$ are non-empty, the equivalence class of a couple has exactly $p^2$ elements, hence we get:
$$ p^2 \mid \binom{2p}{p}-2 $$
that is:
$$ \binom{2p}{p}\equiv 2\!\!\pmod{p^2}.$$
A: Answer that lacks the application of group actions (so it might be useless for you).
For a proof of: $$\binom{2p}{p}=\sum_{k=0}^p\binom{p}{k}^2=2+\sum_{k=1}^{p-1}\binom{p}{k}^2$$
have a look here.
Evidendly prime $p$ divides $\binom{p}{k}$ for each $k\in\{1,\dots,p-1\}$.
I don't know, but maybe some action can be found on a set of cardinality $\binom{2p}{p}$ that is split up in orbits of sizes $\binom{p}{k}^2$
A: Hint: Consider the rotations of a regular polygon with $2p$ vertices, $p$ of which are to be colored black and the others white.  Of the $2p\choose p$ colorings, there are $2$ that alternate black and white.  Can you show that the rest come in groups of $p$?
