divisibility of binomial coefficients by primes

Question

If $p$ is a prime number and $r = 1,2,3 \dots ,p -1$, how can I prove that the binomial coefficients $_nC_r(p,r)$ are divisible by $p$? By divisible I mean $_nC_r(p,r)/p$ leaves remainder $0$.

oivind

Answer 1

Let $p$ be prime. I assume that you are asking why $\binom{p}{r}$ is divisible by $p$ for $r=1,2,\dots,p-1$. Instead of ${}_pC_r$, I will use the notation $\binom{p}{r}$.

Let $N=\binom{p}{r}$. Then $$N=\frac{p!}{r!(p-r)!},\quad \text{or equivalently}\quad p!=Nr!(p-r)!.$$

Clearly $p$ divides $p!$. Thus $p$ divides $Nr!(p-r)!$. But if a prime divides a product, then it divides at least one of the terms. Since $p$ cannot divide $r!$ or $(p-r)!$, it must divide $N$.

Answer 2

Another way: There is an action of $\mathbb Z_p$ on its subsets of size $r$. It's easy to see that if $r$ is as you required, then this action has no fixed points, and, being an action of a $p$-group on a finite set, you get that $\mod p$ the size of the set and number of fixed points are equal. So you have your divisibility relation. Note that using this congruence you can also prove Fermat's Little Theorem, but you can also prove it this way: you can see that $\mathbb Z _p$ acts on the set of functions from a set of $p$ elements to a set of $a$ elements, whose size is $a^p$, and it's easy to see that only the constant functions are the fixed points of this action, and that there are $a$ of these. Then, using the above result again, you have $a^p \equiv a \mod p$.

Edit: Okay, I will explain it in more detail. Take a subset $A$ of $\mathbb Z_p$. Take $j$ in $\mathbb Z_p$, and consider the set $j+A$. This defines an action of $\mathbb Z_p$ on the subsets of $\mathbb Z_p$. Of course, sets of size $r$ are sent to sets of size $r$ by this action. Thus, this defines an action of $\mathbb Z_p$ on its subsets of size $r$. This gives you a decomposition of them into orbits. By the Orbit-Stabilizer Lemma, each orbit has a length that divides $p$. Then the length has to be $1$ or $p$. Since $p \mod p$ is $0$, you can count, $\mod p$, just the fixed points (i.e. orbits of size $1$). But it is clear that a set of size $r$, with $1 < r < p$, cannot be fixed by any translations, and hence there are no fixed points, so the orbits are all of length $p$. Then their sum (which gives you the binomial) is $0 \mod p$.

In a very similar way, you can argue about the following action on functions: given $j$ in $\mathbb Z_p$, and $f$ from $\mathbb Z_p$ to a set of size $a$, you send $f(x)$ to $f(x+j)$. You can check that a function is fixed by such an action exactly when it is a constant function, so for the same reason I have explained above, you have the number of functions = number of constant functions $\mod p$ that translated numerically gives you Fermat's Little Theorem (that can be anyway derived from the above binomial congruence by induction).