Prime $p$ does not divide $\dbinom{n}{p^e}$

Question

I encountered this statement as a lemma in the course of proving the first Sylow theorem.

If $|G|=n=p^em$, $p$ is a prime and $m$ and $p$ are coprime, show that $p$ does not divide $\dbinom{n}{p^e}$.

I tried expanding to factorials, but it didn’t work.

A solution or directive towards the proof is appreciated.

generalization: https://en.wikipedia.org/wiki/Lucas%27s_theorem — anon, Oct 31 '21 at 07:56
This follows from a neat formula for the binomial coefficients modulo a prime known as Lucas's theorem. — Jyrki Lahtonen, Oct 31 '21 at 07:56
Locally, I sketched a proof of Lucas's theorem as a part of this old answer. Probably not the best source on Math.SE, only one of the earliest. It is essentialy one of the alternative proofs in Wikipedia. — Jyrki Lahtonen, Oct 31 '21 at 07:59
Thank you! Lucas theorem is exactly what this is...I think I’ll go on to answer this — USSeR, Oct 31 '21 at 08:17

score 2 · Answer 1 · answered Oct 31 '21 at 09:00

As stated above, this is a consequence of Lucas’ Theorem.

$\mathbf{Lucas’ theorem}:$ Lucas’ Theorem

We have $n=m.p^e$ and $p^e=1.p^e$, which remain the same in base $p$.

According to Lucas’ theorem, $\dbinom{n}{p^e}\equiv \dbinom{m}{1}\pmod p$ .

Since $m>0$, and $m$ and $p$ are coprime, it is implied that $\dbinom{n}{p^e}$ leaves a non-zero remainder on being divided by $p$.

Thus, $p$ does not divide $\dbinom{n}{p^e}$.

score 2 · Answer 2 · answered Oct 31 '21 at 10:05

It seems to me that nothing but the definition of the binomial coefficient is needed for this. Indeed, by the definition of the binomial coefficient we have $$ {n\choose p^e}= \frac{n(n-1)(n-2)\ldots(n-p^e+1)}{1\cdot2\cdots(p^e-1)\cdot p^e}= \frac{n}{p^e}\prod_{k=1}^{p^e-1}\frac{n-k}{k}. $$ It remains to be seen that none of the factors of the last expression is divisible by $p$.

Prime $p$ does not divide $\dbinom{n}{p^e}$

2 Answers2