# Proving Sylow's first theorem

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 $$gS_i = S_i$$?

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

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.

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 $\mid Hs\mid=\mid H\mid\le p^n$.

• 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.
– Huy
Aug 21, 2013 at 15:55
• Well, when $S$ is a subgroup, the notation $\mid S\mid$ means the order of $S$. So you did say the order. :) Aug 21, 2013 at 15:57
• Aren't order and cardinality of sets the same thing?
– Huy
Aug 21, 2013 at 15:59
• 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. :) Aug 21, 2013 at 16:00
• 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.
– Huy
Aug 21, 2013 at 16:01