# Sylow

original sillier post

Note on notation: I’m going to use $$\text{Stab}(x)$$ instead of $$G_x$$ for the stabilizer subgroup and $$\text{Cl}(x)$$ instead of $$^Gx$$ for the conjugacy classes. For the orbit of $$x$$ I’ll stick with the norm and use $$Gx$$, although it’s only used in confusing summation notation that I’ll explain with words too.

We keep using this silly counting argument which I thought was something like Burnside’s lemma but actually is a lot simpler, just partitioning the set into orbits and slapping the orbit-stabilizer theorem on.

If $$G$$ is the group and $$S$$ is the set then

$\vert S\vert = \sum_{Gx} \vert Gx\vert = \sum_{Gx} \frac{\vert G\vert}{\vert\text{Stab}(x)\vert}$

where in the sum, $$x$$ ranges over a set of representatives of the orbits, i.e. exactly one $$x$$ is taken from each orbit, i.e. $$Gx$$ is equal to each orbit exactly once. Anyway, that sum is a terrible abuse of notation so I’ll just leave it out and explain things in English.

Below, let $$p$$ denote some fixed prime.

### Cauchy’s Lemma

Let $$G$$ be a finite abelian group. Then if $$p \mid \vert G\vert$$ then $$G$$ has a subgroup of order $$p$$.

Proof: Induct on the order of $$G$$.

If we have an element $$a$$ with order $$pr$$ for some positive integer $$r$$ then $$\langle a^r \rangle$$ (the subgroup generated by $$a^r$$) works.

Otherwise, we can at least pick an element $$a \neq 1$$ with order not divisible by $$p$$. As $$G$$ is abelian, we can take the factor group $$G/\langle a \rangle$$, whose order is divisible by $$p$$. By induction, this factor group contains a subgroup with order $$p$$; taking a generator of it, we obtain a coset $$b\langle a\rangle$$ with order $$p$$, where $$b \in G$$. Then $$b$$ has order divisible by $$p$$, because if $$b^s = 1$$ then $$(b\langle a\rangle)^s = \langle a\rangle$$, the identity of the factor group, and $$p \mid s$$. As before, let $$b$$’s order be $$pr$$ and take $$\langle b^r \rangle$$.

(Alternatively: this is entirely trivial if you know the structure of all finitely generated abelian groups.)

— end Cauchy’s lemma —

### Sylow I

Let $$G$$ be a finite group, let $$p$$ be a prime and let $$k$$ be a positive integer. If $$p^k \mid \vert G\vert$$ then there exists a subgroup $$H$$ of $$G$$ with order $$p^k$$.

Proof:

Consider the group action $$G$$ acting via conjugation ($$g: x \mapsto gxg^{-1}$$) on the set $$G$$ itself.

Then

$\vert G\vert = \sum \vert\text{Cl}(x)\vert = \sum \frac{\vert G\vert}{\vert C(x)\vert}$

where in the sum, $$x$$ ranges over a set of representatives of the conjugacy classes, or orbits under conjugation.

Consider the center $$C(G)$$. It’s just the set of $$x$$ for which $$\vert C(x)\vert = \vert G\vert$$, so $$\frac{\vert G\vert}{\vert C(x)\vert} = 1$$ and the conjugacy class of $$x$$ contains only $$x$$ itself. We separate them from the summation like so:

$\vert G\vert = \vert C(G)\vert + \sum \frac{\vert G\vert}{\vert C(x)\vert}$

where in the sum, $$x$$ ranges over a set of representatives of the conjugacy classes with at least two elements.

I.1. If $$\vert C(G)\vert$$ is divisible by $$p$$:

Of course, $$C(G)$$ is abelian. So, by Cauchy’s Lemma above, there exists a subgroup $$L < C(G)$$ with order $$p$$. As $$L$$ is a subgroup of the center, it commutes with every element of $$G$$, so it’s normal. Then we can compute the factor group $$G/L$$, whose order is divisible by $$p^{k-1}$$. Thus by induction $$G/L$$ contains a subgroup $$H$$ with order $$p^{k-1}$$. The union of all the cosets in $$H$$ is a subgroup with order $$p^k$$, so we’re done.

I.2. If $$\vert C(G)\vert$$ is not divisible by $$p$$:

Since $$\vert G\vert$$ is divisible by $$p$$, there would have to be another term on the RHS of the equation not divisible by $$p$$. Thus, for some $$x \in G$$, the value $$\frac{\vert G\vert}{\vert C(x)\vert}$$ is not divisible by $$p$$. This is only possible if $$p^k \mid \vert C(x)\vert$$. But $$C(x) \lneqq G$$ (if the equality sign held $$x$$ should have been moved out into $$C(G)$$), hence by the induction hypothesis on $$C(x)$$ we have a subgroup $$L < C(X) < G$$ with $$\vert L\vert = p^k$$ and we’re done.

— end Sylow I —

### Sylow II

Let $$G$$ be a finite group; suppose $$p^k \parallel \vert G\vert$$ (i.e. $$p^k \mid \vert G\vert, p^{k+1} \nmid \vert G\vert$$). Consider the collection $$\mathcal{S} = \{H < G \mid \vert H\vert = p^k \}$$ of all subgroups of $$G$$ with order $$p^k$$ (which are called Sylow subgroups).

Then:

1. for any two Sylow subgroups $$H_1, H_2 \in \mathcal{S}$$, they are conjugate, i.e. $$\exists x \in G$$ such that $$xH_1x^{-1} = H_2$$;
2. $$\vert\mathcal{S}\vert$$ is a divisor of $$\frac{\vert G\vert}{p^k}$$, the index of any Sylow subgroup;
3. $$\vert\mathcal{S}\vert\equiv 1 \bmod p$$;
4. for any subgroup $$L < G$$ with order $$p^r$$ for some nonnegative integer $$r$$, there is some Sylow subgroup $$H \in \mathcal{S}$$ with $$L < H$$.

Proof.

First, we prove this statement: (d’) given a fixed Sylow subgroup $$H$$, for any subgroup $$L < G$$ with order $$p^r$$ for some nonnegative integer $$r$$, there exists a conjugation $$xHx^{-1}$$ of $$H$$ such that $$L < xHx^{-1}$$.

Consider the group action $$L$$ acting via left translaion ($$\ell: gH \mapsto \ell gH$$) on the set of left cosets of $$H$$, which we’ll denote $$\mathcal{T} = \{gH \mid g \in G\}$$.

Again:

$\vert\mathcal{T}\vert = \sum \frac{\vert L\vert}{\vert\text{Stab}(gH)\vert}$

where $$gH$$ ranges over a set of representatives of the orbits in $$\mathcal{T}$$.

Now, we certainly know that $$\vert\mathcal{T}\vert = \frac{\vert G\vert}{\vert H\vert = p^k}$$ is not divisible by $$p$$. Hence again there’s a term on the RHS that’s not divisible by $$p$$. As $$\vert L\vert = p^r$$ this is only possible if, for some coset $$gH$$, we have $$\text{Stab}(gH) = L$$.

Thus, that coset $$gH$$ satisfies: for any $$\ell \in L$$, we have $$\ell gH = gH$$, i.e. $$\ell \in gHg^{-1}$$. Then $$L \subset gHg^{-1}$$, and as both are subgroups of $$G$$ we’re done.

With this we have proved (d), since (by Sylow I) at least one $$H$$ exists, and the $$gHg^{-1}$$ produced is a subgroup with the same order as $$H$$ and is thus another Sylow subgroup. It also implies (a), just by setting $$L$$ to be another Sylow group.

Now we prove (c). Fix a Sylow subgroup $$H_1$$ and consider the group action $$H_1$$ acting via conjugation ($$x: H \mapsto xHx^{-1}$$) on the set of Sylow subgroups $$\mathcal{S}$$. Then

$\vert\mathcal{S}\vert = \sum \frac{\vert H_1\vert}{\vert\text{Stab}(H_j)\vert}$

where $$H_j$$ ranges over a set of representatives of the orbits in $$\mathcal{S}$$.

If $$H_j = H_1$$ it’s obvious that every element of $$H_1$$ is in the stabilizer, hence the summand is $$1$$.

On the other hand, if $$H_j \neq H_1$$, we claim that $$\text{Stab}(H_j) \neq H_1$$. Suppose otherwise; then $$xH_jx^{-1} = H_j$$ for any $$x \in H_1$$, so $$H_j$$ is a normal subgroup of the subgroup generated by these two Sylow subgroups, $$M = \langle H_1, H_j \rangle$$.

Consider the factor group $$M/H_j$$; we claim that its order is also a power of $$p$$. This is because if we take the intersection of each coset with $$H_1$$, we actually get a factor group of $$H_1$$. (I think it’s the “first isomorphism theorem” or something.) That is, $$M/H_j \cong H_1/(H_1 \cap H_j)$$ so its order is a divisor of $$\vert H_1\vert = p^k$$.

Thus $$\vert M\vert = p^k\vert M/H_j\vert$$ is also a power of $$p$$, and $$\vert M\vert > \vert H_1\vert = p^k$$. But as $$M$$ is a subgroup of $$G$$, we know $$\vert M\vert$$ (a higher power of $$p$$ than $$p^k$$) divides $$\vert G\vert$$, contradicting our selection of $$k$$.

Therefore, $$\text{Stab}(H_j) \neq H_1$$ and therefore $$\frac{\vert H_1\vert}{\vert\text{Stab}(H_j)\vert}$$ is greater than 1. As $$\vert H_1\vert = p^k$$, the result is divisible by $$p$$. Therefore, all terms vanish $$\bmod p$$ except for the orbit of $$H_1$$ whose summand is $$1$$, so $$\vert\mathcal{S}\vert\equiv 1 \bmod p$$.

Now, (b) is easy. Consider the group action $$G$$ acting via conjugation ($$g: H \mapsto gHg^{-1}$$) on the set of Sylow subgroups $$\mathcal{S}$$. Since there’s only one orbit,

$\vert\mathcal{S}\vert = \frac{\vert G\vert}{\vert\text{Stab}(H)\vert}$

from which it’s obvious $$\vert G\vert$$ is divisible by $$\vert\mathcal{S}\vert$$. And since $$\vert\mathcal{S}\vert$$ is not divisible by $$p$$, we know that $$\vert G\vert /p^k$$ is divisible by $$\vert\mathcal{S}\vert$$ too.

— end Sylow II —

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