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 \(p^k\).


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


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

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).


  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 \(L < H\).


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 \(\mathcal{T} = \{gH \mid g \in G\}\).


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

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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