Lemma for Sylow-Theorem

Question

As a start to proof the Sylow theorems, we proved the following lemma. I would like to understand the proof, I know there might be other proofs but I would appreciate your help with this particular one. Let $G$ be a finite group, $p$ prime so that ord$(G)=n=p^lm$ with $l\geq 1, p\nmid m$. I want to show that $$ p\nmid{n\choose p^l} $$

The proof takes $$ {n\choose p^l}=\frac{n(n-1)\dotsm(n-r)\dotsm(n-p^l+1)}{p^l(p^l-1)\dotsm(p^l-r)\dotsm.1} $$ and looks at specific $$ \frac{(n-r)}{(p^l-r)}, \, 0\leq r \leq p^l-1 $$ so at $$\frac{n}{p^l},\frac{(n-1)}{(p^l-1)},...,\frac{(n-(p^l-1))}{(p^l-(p^l-1))}=\frac{(n-p^l+1)}{1}$$ We can write $r=p^ks$ with $p\nmid s$.

The proof now states the following, which I don't understand:

(i) $k\leq l-1$ But what if $r=0$? Then $s=0$ and $k$ can be everything, also smaller than $l-1$ or not? Is this always true?

(ii) $k$ is the biggest power contained in $n-r$. But again: What if $r=0$? I think I have the most trouble understanding why $k\leq l-1$.

(ii) $p^k\mid(p^l-r)$ I understand how this follows. But why does this show the lemma in the end?

Thank you in advance for your help.

Answer 1

Lemma. Let $m$ and $k$ be positive integers and $p$ a prime such that $p \nmid m$. Then:

$$p \nmid \binom{mp^k}{p^k}$$

Proof. We get:

\begin{alignat}{1} \binom{mp^k}{p^k} &= \frac{(mp^k)!}{p^k!(mp^k-p^k)!} \\ &=\frac{mp^k(mp^k-1)\dots (mp^k-i)\dots(mp^k -(p^k-1))(mp^k-(p^k-1)-1)!}{p^k!(mp^k-p^k)!} \\ &= \frac{mp^k(mp^k-1)\dots (mp^k-i)\dots(mp^k-(p^k-1))(mp^k-p^k)!}{p^k!(mp^k-p^k)!} \\ &= \frac{mp^k(mp^k-1)\dots (mp^k-i)\dots(mp^k-(p^k-1))}{p^k!} \\ &= \frac{m(mp^k-1)\dots (mp^k-i)\dots(mp^k-(p^k-1))}{(p^k-1)!} \\ &= m\prod_{i=1}^{p^k-1}\frac{(mp^k-i)}{(p^k-i)} \\ \tag 1 \end{alignat}

Now, $p \mid \prod_{i=1}^{p^k-1}\frac{(mp^k-i)}{(p^k-i)} \Rightarrow \exists \tilde i, 1 \le \tilde i\le p^k-1$ such that $\tilde i=lp^j$, with $j \ge 1$ and $p \nmid l$; but $p \nmid \frac{mp^k-\tilde i}{p^k-\tilde i}$, because either $\frac{mp^k-\tilde i}{p^k -\tilde i}=\frac{mp^{k-j}-l}{p^{k-j}-l}$ (for $j\le k$) or $\frac{mp^k-\tilde i}{p^k -\tilde i}=\frac{m-lp^{j-k}}{1-lp^{j-k}}$ (for $j > k$), and both $p \nmid l$ and $p \nmid m$. So, $p$ doesn't divide any factor in the product and, by $(1)$, it doesn't divide $\binom{mp^k}{p^k}$ either. $\Box$

Answer 2

Consider first the case $r > 0$.

First, where does the bound on $k$ come from? Since $r < p^l -1$, it cannot be divided by a power $p^k$ with $k \ge l$ (An integer cannot be divided by an integer of greater absolute value!). Therefore if we write $r = p^k s$, with $s$ another integer, we must have $k \le l-1$.

$p^k\ |\ (p^l-r)$ by the construction of $k$. Since $p^k$ is also the greatest power of $p$ dividing $(n-r)$, none of the fractions $\frac{n-r}{p^l-r}$ contain positive powers of $p$.

For $r=0$, note that $\frac{n}{p^l} = m$ which is also not divided by $p$. So, none of the factors of $n\choose p^l$ are divided by positive powers of $p$ and as $p$ is prime this completes the proof.