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So, I'm having some problems with understanding one part of the proof of Sylow's Theorem.

Let $G$ be a group of order $n$=$p^{a}m$, where $p$ is prime and doesn't divide $m$. Then: (a) $G$ has a subgroup of order $p^{a}$....

So, here is the proof I found in Cameron's Permutation groups book:

To show (a), we let $\Omega$ be the set of all subsets of $G$ of cardinality $p^{a}$. Then $\Omega$ is a set of cardinality $\binom{n}{p^a}$. We define an action of $G$ on $\Omega$ by right multiplication: $\mu(S,g)=Sg$ for $S\in\Omega$ $g\in G$. Then $\Omega$ is the union of the $G$-orbits of this action. Note that each $G$-orbit, as a set of subsets of $G$, covers all of $G$: for if $x\in S$, then $y\in Sx^{-1}y$. So there are two types of orbit: Type 1:orbits containing exactly $n/p^{a}=m$ sets, covering $G$ without overlapping; Type 2:orbits containing more than $m$ sets. Now if $S$ is in Type 1 orbit, then its stabiliser has order $n/m=p^a$, and so is a subgroup $P$ of the required order.

The next part is a problem for me:

On the other hand, if $S$ is in a Type 2 orbit $\Delta$, then $|\Delta|>m$, so $|\Delta|$ is divisible by $p$. So, if we can show that $|\Omega|$ is not divisible by $p$, then it will follow that there must be Type 1 orbits, and (a) will be proved.

So, first, I don't get why is $|\Delta|$ divisible by $p$. Actually I don't get nothing from that problematic part so I would be thankful if someone could give me a hint or something.

Thanks!!

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What is the largest divisor of $n$ which is not divisible by $p$? (use the factorisation) –  Mark Bennet Dec 9 '11 at 17:45

2 Answers 2

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The orbit-stabilizer theorem tells us that, if $\Delta$ is the orbit of some $x\in \Omega$, then $$|G|=p^am=|\Delta|\cdot|\operatorname{stab}(x)|.$$

For any natural number $k$, if $\gcd(k,p)=1$, then $k\mid p^am$ implies that $k\mid m$. Thus, if $\gcd(|\Delta|,p)=1$, then $|\Delta|\;\;\big\vert\;\; m$. But if $|\Delta|>m$, then we must have $|\Delta|\nmid m$. Thus $\gcd(|\Delta|,p)=p$, i.e $p\mid |\Delta|$.

Here's what's going on in the next sentence of the problematic part: we know that the orbits $\Delta_i$ of the action of $G$ on $\Omega$ partition $\Omega$. Thus, we have that $$|\Omega|=\sum_i|\Delta_i|$$ If every orbit $\Delta_i$ were of Type 2, then every term in the sum would be divisible by $p$, and therefore so also would be $|\Omega|$. Thus, if we prove that $|\Omega|$ is not divisible by $p$, there is at least one orbit $\Delta_i$ for which $|\Delta_i|$ is not divisible by $p$, and this orbit will necessarily be of Type 1.

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Thanks! I got it.. –  aldo Dec 9 '11 at 18:04
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Glad to help! If you've found my answer and Brian's answer useful, you can upvote them, and accept one if you like. –  Zev Chonoles Dec 9 '11 at 18:07

The key is that $|\Delta|$ must divide $n$, the order of $G$. Since $n=p^am$, and $p\nmid m$, every divisor of $n$ that is bigger than $m$ must include a factor of $p$ by the unique factorization theorem.

Let $H=\{x\in G:Sx=S\}$, the stabilizer of $S$. $H$ is a subgroup of $G$, and it’s easy to see that for any $x,y\in G$, $Sx=Sy$ iff $yx^{-1}\in H$ iff $y\in Hx$, so the members of the orbit $\Delta$ correspond bijectively to the right cosets of $H$. There are $n/|H|$ of these, so $|\Delta|=n/|H|$, and hence $$\frac{n}{|\Delta|}=|H|\;,$$ i.e., $|\Delta|$ divides $n$.

Added:

$\Omega$ is the union of the orbits, so if $p$ does not divide $|\Omega|$, it’s impossible for all of the orbits to have sizes divisible by $p$, and therefore they can’t all be of Type 2.

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