Cardinality of a p-Sylow

Question

Let $G$ be a group with cardinal $n=p^{\alpha}m$, (prime $p$ dividing $n$ and $\gcd(m,p)=1$). Let $E$ be the set of subsets of $G$ containing $p^{\alpha}$ elements. I'm trying to understand why $p$ does not divide $\vert E \vert = \binom{p^{\alpha}m}{p^{\alpha}}$

Writing

\begin{equation} \displaystyle \binom{p^{\alpha}m}{p^{\alpha}}=\frac{p^{\alpha}m}{p^{\alpha}}.\frac{p^{\alpha}m-1}{p^{\alpha}-1}...\frac{p^{\alpha}m-p^{\alpha}+1}{1} \end{equation}

After simplifications, the remaining quantity is not divisible by $p$.

The first fraction of the binomial coefficient can be simplified by $p^{\alpha}$ but is it true for the other fractions ?

I thank you in advance for any suggestions.

Answer 1

Yes, it's true for the other fractions as well. Let $k\in\{1,2,...,p^{\alpha}-1\}$. We can write $k=p^sb$ where $p$ does not divide $b$. Obviously $0\leq s\leq \alpha-1$. Then:

$p^{\alpha}m-k=p^{\alpha}m-p^sb=p^s(p^{\alpha-s}m-b)$

The expression in the brackets is not divisible by $p$, hence $p$ divides $p^{\alpha}m-k$ with multiplicity $s$. Similarly:

$p^{\alpha}-k=p^{\alpha}-p^sb=p^s(p^{\alpha-s}-b)$

So again, $p$ divides $p^{\alpha}-k$ with multiplicty $s$.

Answer 2

Recall that the $p$-adic valuation of $N$ is $v_p(N)=r$ when $p^r$ is the highest power of $p$ dividing $N$, so that $(p,N)=1$ if and only if $v_p(N)=0$.

By counting powers of $p$ smaller than $n$ we get the formula $$ v_p(n!)=\lfloor \frac np\rfloor+\lfloor\frac n{p^2}\rfloor+\lfloor\frac n{p^3}\rfloor+\cdots. $$ Now let $n=p^am$ with $(p,m)=1$. The formula above reads $$ v_p((p^am)!)=p^{a-1}m+p^{a-2}m+\cdots +m. $$ Using the same formula again we get $$ v_p((p^a(m-1))!)=p^{a-1}(m-1)+p^{a-2}(m-1)+\cdots +(m-1) $$ and $$ v_p((p^a!)=p^{a-1}+p^{a-2}+\cdots+p. $$ Now since $v_p(\frac A{BC})=v_p(A)-v_p(B)-v_p(C)$ we can apply the above computation to $$ v_p(\binom{p^am}{p^a})=v_p((p^am)!)-v^p(p^a!)-v_p((p^a(m-1))!)=0 $$ proving the assertion.

Answer 3

Take general term $\frac{p^am-i}{p^a-i}$. Here $i$ is between $\{0,1,\cdots,p^a-1\}$.

The number $i$ may or may not be divisible by $p$; however, we can always write $i=p^bt$ where $0\le b< a$ and $p\nmid t$.

Now the general term becomes $\frac{p^am-p^bt}{p^a-p^bt}$. You can see that $p^b$ cancels and then the numerator and denominator is not divisible by $p$.

Upshot: If, in general term, the numerator is divisible by $p^b$ then so is the denominator and vice-versa. Hence your conclusion!