Combinatoric proof $p^2$ divides combination ${2p\choose p}-2$. Give a combinatoric proof for this problem for primary $p$:
$$
p^2 \left| \Big(\begin{array}{cc}2p\\p\end{array}\Big)-2\right. 
$$
 A: Let $\mathcal{X}=\{0,1\}\times\{1,2,\cdots,p\}$ and let $\mathcal{S}$ be the collection of all $p$-subsets of $\mathcal{X}$, except for the two subsets $\{0\}\times\{1,\cdots,p\}$ and $\{1\}\times\{1,\cdots,p\}$.
Interpret the second coordinate as an integer mod $p$. Consider the functions $f_{k,l}$ defined by
$$ f_{k,l}(x,y)=\begin{cases} (x,y+k) & x=0 \\ (x,y+l) & x=1 \end{cases} $$
Given any $A\in\mathcal{S}$, we have $f_{k,l}(A)=A$ if and only if $(k,l)=(0,0)$. Therefore we partition $\mathcal{S}$ into groups of $p^2$ members, where any two $A,B\in\mathcal{S}$ are in the same group if and only if $B=f_{k,l}(A)$ for some $(k,l)$. (Exercise: why does each group have $p^2$ members?)

Here is the advanced view of the above argument. If $G$ acts on $X$ and $H$ acts on $Y$ then there is a canonical induced action of $G\times H$ on $X\sqcup Y$. Thus, $\mathbb{Z}_p^2$ acts on $\{0,1\}\times\mathbb{Z}_p$. If we use the choose notation $\binom{X}{k}$ to represent the collection of $k$-subsets of $X$, then 
$$\mathbb{Z}_p^2 ~\textrm{ acts freely on }~ \binom{\{0,1\}\times\mathbb{Z}_p}{p}-\left\{ \begin{array}{c} \{0\}\times\mathbb{Z}_p, \\ \{1\}\times\mathbb{Z}_p \end{array} \right\}.$$
If $G$ acts freely on $X$ then $X\cong G\sqcup\cdots\sqcup G$ as $G$-sets, and in particular $|G|$ divides $|X|$.
