1

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

Jack D'Aurizio
  • 353,855
HaroldF
  • 1,520

3 Answers3

2

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: $$ \binom{2p}{p}\equiv 2\!\!\pmod{p^2}.$$

Jack D'Aurizio
  • 353,855
  • Thank you, so I have some questions:
    1. The fact that I look at S as a couple of that type is only to see if S which between 1,p,p+1 and 2p is contained in S?

    2)What is equivalence class in this case means? The orbit of an element? 3)Why that "iff" holds?

     – HaroldF Oct 31 '15 at 15:15
  • @HaroldF: 1) Exactly, we split a subset into its "left part" an "right part" 2) $C_p$ acts on the left part by sending an element into its consecutive element, but $p$ into $1$; the same happens when $C_p$ acts on the right part: the action of $C_p$ maps $p+1$ into $p+2$ and so on, till $2p$ that is mapped to $p+1$. So we have that every subset, having non-empty left part and non-empty right part, can be mapped into $p^2$ different subsets by the action of $C_p\times C_p$. That gives an equivalence relation on the set of subsets having non-empty left and right parts, – Jack D'Aurizio Oct 31 '15 at 15:24
  • in which every equivalence class has $p^2$ elements. It follows that $\binom{2p}{p}-2\equiv 2!!\pmod{p^2}$ as stated. – Jack D'Aurizio Oct 31 '15 at 15:25
  • Thank you, now i see the point! – HaroldF Oct 31 '15 at 15:48
1

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 $\binom{p}{k}^2$

Community
  • 1
drhab
  • 151,093
1

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 $p$?

Barry Cipra
  • 79,832
  • Thank you, like this point of view, but I cannot see the point, can you be more specific? – HaroldF Oct 31 '15 at 15:48
  • @HaroldF If the $\binom{2p}{p}$ colorings are $2$ plus a bunch of groups of $p$, then what is $\binom{2p}{p}$ mod $p$? There is no point in considering $C_p\times C_p$ here, $C_p$ is much more natural. This is also the point of Weaam's comment above, the linked wikipedia section which you should read. – anon Nov 02 '15 at 07:44
  • Yes, my question to the Point of Barry is that I cannot see why other colourings come in groups of p – HaroldF Nov 02 '15 at 08:20
  • @HaroldF The orbits all have size dividing $C_p$, so either $1$ or $p$. There are two fixed points (two rotation-invariant colorings), so the rest come in sets of $p$ (namely, in orbits of size $p$). – anon Nov 03 '15 at 03:12