Cauchy's Theorem and Sylow Theorems

(Cauchy's Theorem) For a finite group \(G\) and a prime number \(p\) that divides the order of \(G\) (the number of element in \(G\)), then there exists an element of order \(p\) in \(G\). That is, an element \(x\) such that the smallest positive integer \(n\) with \(x^n=e\) is \(p\).

Proof:

We define a set

\[ S=\{(x_1,\dots,x_p)\in G^p|x_1x_2\cdots x_p=e\} \]

We see that any element in \(S\) is uniquely determined by the first \(p-1\) components, since

\[ x_p=(x_1x_2\cdots x_{p-1})^{-1} \]

Thus,

\[ |S|=|G|\times |G|\times\cdots\times |G|=|G|^{p-1} \]

Moreover, notice that \(aa^{-1}=a^{-1}a=e\) for all \(a\in G\). For \((x_1,x_2,\dots,x_p)\) in \(S\),

\[ x_2x_3\cdots x_p x_1=x_1^{-1}x_1=e\implies (x_2,x_3,\dots,x_p,x_1)\in S \]

We can see this as cyclicly permuting the components of \(x\in S\). By the orbit-stabilizer Theorem, orbits in \(S\) under this action have size \(1\) or \(p\). The former one only occurs for tuples \((x,x,\dots,x)\in S\) such that \(x^p=e\).

We count the number of elements of \(S\) by orbits, then we know that the number of \(x\in G\) such that \(x^p=e\) has to be divisble by \(p\). Since \(x=e\) is one such element, we know that there must be at least \(p-1\) other elements as solutions for \(x\). They all have order \(p\) and this completes the proof.

(Sylow Theorems)

1, A finite group \(G\) whose order \(|G|\) is divisible by \(p^k\) has a subgroup of order \(p^k\).

We can apply a similar idea to prove this theorem.

Proof:

Let \(|G|=p^km=p^{k+r}u\) such that \(p\nmid u\). Let \(S\) denote the set of subsets of \(G\) with size \(p^k\). \(G\) acts on \(S\) by left multiplication, for \(g\in G\) and \(s\in S\),

\[ g\cdot s=\{gx|x\in s\} \]

For a given set \(s\in S\), define its stabilizer subgroup

\[ G_s=\{g\in G|g\cdot s = s\} \]

and its orbit

\[ Gs=\{g\cdot s|g\in G\} \]

It suffices to show that there exist some \(s\) such that \(|G_s|=p^k\) since \(G_s\) is a subgroup of \(G\). This is the maximal possible size of \(G_s\), since for any fixed element \(\alpha\in s\subseteq G\), the right coset \(G_s\alpha\) is a subset of \(s\). Hence

\[ |G_s|=|G_s\alpha|\le |s|=p^k \]

By the orbit-stabilizer theorem, we have

\[ |G_s||Gs| = |G|=p^{k+r}u \]

for all \(s\in S\). Define \(v_p(n)\) as the largest non-negative integer \(k\) such that \(p^k|n\). If we count the number of factors of \(p\) on both sides, we get

\[ v_p(|G_s|) + v_p(|Gs|) = v_p(|G|)=k+ r \]

This implies that when \(v_p(|G_s|)< k\) we have \(v_p(|Gs|) > k+r-k=r\). Notice that in order to have \(v_p(|G_s|)=k\), one would have \(|G_s|=p^k\), which are the ones we are looking for.

Now we count the number of elements in \(S\) in two ways. By basic combinatorics, we know that

\[ |S|=\binom{p^km}{p^k} \]

and \(|S|\) is also the sum of \(|Gs|\) over all distinct orbits \(Gs\). By Kummer's theorem we get

\[ v_p(|S|)=r \]

If we don't have any \(s\) such that \(|G_s|=p^k\), then all \(s\in S\) have orbit \(Gs\) with \(v_p(|Gs|) > r\). The sum of them contradicts with \(v_p(|S|)=r\), completing the proof!