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Let $p$ be prime and $G$ a group such that $|G| = p^n k$, where $p \nmid k$. We want to show that $G$ has at least one Sylow $p$-subgroup.

Let $\mathcal{S} := \{ S \subset G \mid \, |S| = p^n\}, |\mathcal{S}| := N$. Then, $N = \binom{p^n k}{p^n} \equiv k \text{ (mod } p)$. Let $G$ act on $\mathcal{S}$ by $(g, S) \mapsto gS = \{gs\mid \, s \in S\}$. Let $S_1, \dots, S_r$ represent the disjoint orbits of $\mathcal{S}$. Then, $p$ cannot be a divisor of the cardinality of all orbits, because $N \equiv k \text{ (mod } p)$. Choose $S_i$ such that $p \nmid |GS_i|$. Then, $\operatorname{Stab}_G(S_i) = S_i$ and thus $S_i$ is a Sylow $p$-subgroup.

Why is $\operatorname{Stab}_G(S_i) = S$? We defined the stabilizer as follows:

$$\operatorname{Stab}_G(x) := \{g \in G\mid\, gx = x\}$$

Maybe I'm just confusing things, but if we take $g \in G \setminus S_i$, isn't still $gS_i = S_i$?

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  • $\begingroup$ Something seems off. $S_i$ is an orbit, not an element of $\mathcal{S}$. What is the stabilizer of an orbit? $\endgroup$
    – RghtHndSd
    Aug 21, 2013 at 15:48
  • $\begingroup$ $S_i$ only represents the orbit $GS_i$. $\endgroup$
    – Huy
    Aug 21, 2013 at 15:50
  • $\begingroup$ Ah, yes! Sorry, misread the proof. $\endgroup$
    – RghtHndSd
    Aug 21, 2013 at 15:51
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    $\begingroup$ It does not follow that ${\rm Stab}_G(S_i)=S_i$. BTW can you cite your source? $\endgroup$
    – anon
    Aug 21, 2013 at 16:12

2 Answers 2

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You are confusing yourself by the notation. Also observe that the proof requires the notion of stabilizers of sets not points. But let's use your notation. Let $T \in S_i$, where the cardinality of orbit $S_i$ is not divisible by $p$. Put $H=Stab_G(T) = \{g \in G | gT=T\}$. We are going to argue that $H$ is a Sylow $p$-subgroup.
First, since $p$ does not divide $|S_i| = [G:H]$, it follows that $p^n | |H|$, hence $p^n \leq |H|$. Since $H$ stabilizes $T$ by left multiplication, for any $t \in T$ we have $Ht \subseteq T$. Hence $|H| = |Ht| \leq |T| = p^n$. We conclude that $|H| = p^n$ and we are done.

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I surmise that you took the proof in the wrong direction: we do not show that the stabiliser equality, but show that the stabiliser is a sub-group with the correct order.
I think also you meant to say that $p$ cannot divide the cardinality of the orbit of $S_i$, not the order of $S_i$, where $S_i$ is a subgroup in that orbit. Then let the orbit containing $S_i$ be $\mathscr O$.
So $|\mathscr O|=\mid G\mid/\mid H\mid$, where $H$ is the stabiliser of $S_i$.
Thus all the powers of $p$ appearing in the factorisation of $\mid G\mid$ must divide the order of $H$. This is equivalent with saying that $p\mid |H|$.
If the proof is fine in the direction described, tell me, for I cannot see how your proof is going to end. Regards.
Edit The above proof is not complete: we also have to show that the cardinality of $H$ is exactly $p^n$, not just divisible by $p^n$.
Since $H$ stabilises $S_i$, we know that, for every $s\in S_i$, $Hs\subset S$, and hence $\mid Hs\mid=\mid H\mid\le p^n$.

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  • $\begingroup$ Yes, I meant that $p$ cannot divide the cardinality of the orbit $S_i$ (I never said order, did I?). Also, I fixed a mistake. The way I understand the proof is that we have an orbit such that $p$ doesn't divide its cardinality and then find $\operatorname{Stab}_G(S_i) = S_i$ (but I don't know why this equation holds). This equation implies that the stabilizer and $S_i$ have the same cardinality and as the stabilizer is a subgroup (of cardinality $p^n$), we are done. $\endgroup$
    – Huy
    Aug 21, 2013 at 15:55
  • $\begingroup$ Well, when $S$ is a subgroup, the notation $\mid S\mid$ means the order of $S$. So you did say the order. :) $\endgroup$
    – awllower
    Aug 21, 2013 at 15:57
  • $\begingroup$ Aren't order and cardinality of sets the same thing? $\endgroup$
    – Huy
    Aug 21, 2013 at 15:59
  • $\begingroup$ Also, $S_i$ is not necessarily a subgroup! I do not see how can this line of proof work? And yes, the two words are the same, so do not mind that problem. :) $\endgroup$
    – awllower
    Aug 21, 2013 at 16:00
  • $\begingroup$ Yes, $S_i$ is not necessarily a subgroup. That is the point of the equation $\operatorname{Stab}_G(S_i) = S_i$: As the stabilizer IS a subgroup, it implies that $S_i$ is too. $\endgroup$
    – Huy
    Aug 21, 2013 at 16:01

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