# Wielandt's proof of Sylow's Theorem

2017-04-28 ( 2017-04-28)## Wieldandt's proof of Sylow's Theorem

A brief preparatory combinatorial discussion:

[ \binom{n}{k} = \dfrac{n!}{k!(n-k)!} = \dfrac{n (n-1) \ldots (n-j) \ldots (n-k+1)}{ 1 \ldots j \ldots (k-1) \cdot k } = \dfrac{\prod_ {j=1}^k (n-j+1) }{ \prod_ {j=1}^k j } ] and therefore [ \binom{n}{k} = \dfrac{n}{k} \cdot \dfrac{n-1}{1} \ldots \dfrac{n-j}{j} \ldots \dfrac{n-(k-1)}{k-1} = \dfrac{n}{k} \prod_{j=1}^{k-1} \dfrac{n-j}{j} ]

Let $p$ be a prime number. Now, if $n$ is an integer divisible by a power of $p$, say $p^\alpha$, then $n=p^\alpha m$. Let $k=p^\alpha$. Then by the previous equation [ \binom{n}{k} = \binom{p^\alpha m}{p^\alpha} = \dfrac{p^\alpha m}{p^\alpha } \cdot \dfrac{p^\alpha m-1}{1} \ldots \dfrac{p^\alpha m-j}{j} \ldots \dfrac{p^\alpha m -(p^\alpha-1)}{p^\alpha-1} ]

Consider all the fractions $\dfrac{p^\alpha -j}{j}$, for $j=1\ldots p^\alpha -1$: if a power $p^r$ divides $j$, then $p^r \leq j < p^\alpha$ and so $p^r$ divides $p^\alpha$ and therefore $p^r$ divides $p^\alpha m - j$. On the other hand, if a power $p^r$ divides $p^\alpha m - j$, then it cannot be $p^r>p^\alpha$, since $j<p^\alpha$ (exercise?), and hence $p^r\leq p^\alpha$, whence the fact that $p^r|j$ follows. Hence, all powers of $p$ in all fractions cancel out, and the only place where they can appear is in $m$, therefore the following lemma holds.

- (1)
- If $p$ is a prime, $m$ a positive integer, then [ p^r | m \iff p^r | \binom{p^\alpha m}{p^\alpha}. ]

Now, as the proof of Sylow's theorem.
Let $G$ be a group of order $n$, and $p^\alpha$ divides $n$ for a prime $p$.
Let $R_ 1$, $R_ 2$, $\ldots$, $R_ N$ be all the subsets of $G$ of cardinality $p^\alpha$.
There are $N=\binom{n}{p^\alpha} = \binom{p^\alpha m}{p^ \alpha}$ such subsets.
Let $\rho$ denotes the maximum power of $p$ dividing $m$, and hence the maximum power of $p$
dividing $N$: $p^\rho | N$, $p^{\rho+1}\not| N$ because of **(1)**.

Let $G$ act on the set $X$ of all $R_ j$'s by left multiplication: there must be at least one $G$-orbit in $X$ such that the number $\mu$ of its elements is not divisible by $p^{\rho +1}$, since otherwise $N$ would be divisible by $p^{\rho +1}$. Let $R\in X$ be one of the elements in such orbit.

The stabilizer of $R$ is the subgroup $H$ of all elements in $G$ such that $gR=R$.

First, note that for each $h\in H$, $h\neq 1$ induces a fixed-point-free bijection $h\from R\to R$ (which sends $g\in R$ to $hg\in R$). Hence $H$ acts freely on $R$: it follows that $R$ must have at least $|H|$ elements, or equivalently that $|H| \leq p^\alpha$ since $R$ has $p^\alpha$ elements by definition.

Moreover, $H$ is a subgroup of $G$ of index equal the length of the orbit of $R$, that is $\mu$, and hence $|H| = \frac{n}{\mu} = \dfrac{p^\alpha m}{\mu}$. By construction [ p^r | m \implies p^{\alpha + \rho} | p^\alpha m = \mu |H|. ] Since $p^{\rho+1} \not| \mu$ by definition, it must be that $p^\alpha$ divides $|H|$, and hence that $p^\alpha \leq |H|$.

This implies that $|H|=p^\alpha$, and therefore the following theorem holds (it is a version of Sylow's Theorem).

- (S)
- If $p$ is a prime number and $p^\alpha$ divides the order of a finite group $G$, then there exists a subgroup $H\subset G$ of order $p^\alpha$.