Proof of Cauchy's Theorem in Group Theory 📂Abstract Algebra

Proof of Cauchy's Theorem in Group Theory

Theorem¹

For a finite group $G$, if a prime $p$ is a divisor of $|G|$, then there exists a subgroup $H \leqslant G$ satisfying $|H| = p$.

Explanation

When one speaks of Cauchy’s theorem, this theorem is usually not the one that comes to mind. Another Cauchy’s theorem is important enough to form the very foundation of complex analysis, whereas this theorem is rarely mentioned. Above all, since it is generalized by the First Sylow theorem, cases where one actually has to use Cauchy’s theorem are extremely rare.

Whether knowing it will be of any help is uncertain, but the method of proof is remarkably unique in many ways, starting from its very setup. It is recommended to prove it yourself at least once, even if only out of curiosity.

Proof

For $i=1 , \cdots p$, let $g_{i} \in G$ and let the identity of $G$ be $e$. Consider the set $$ X := \left\{ (g_{1} , \cdots , g_{p}) \ | \ g_{1} \cdots g_{p} = e \right\} $$ and the symmetric group $S_{p}$.

$\rho_{1} \in S_{p}$ is a permutation that shifts the tuples of $X$ by one place, performing an action such as $$ \rho_{1} (g_{1}, g_{2} , \cdots , g_{p-1} , g_{p}) = (g_{2}, g_{3} , \cdots , g_{p} , g_{1}) $$ By the definition, $g_{p}$ is determined as $g_{p} = (g_{1} \cdots g_{p-1} )^{-1}$, so $|X| = |G|^{p-1}$, and since $p$ is a divisor of $|G|$, $p$ is a divisor of $|X|$.

Property of $p$-groups: If a finite group $G$ is a $p$-group and $X$ is a $G$-set, then $|X| \equiv |X_{G}| \pmod{p}$

Since $| \left< \rho_{1} \right> | = p$, $$ |X| \equiv \left| X_{ \left< \rho_{1} \right> } \right| \pmod{p} $$ holds, and $p$ is also a divisor of $\left| X_{ \left< \rho_{1} \right> } \right|$. This means that in $\left| X_{ \left< \rho_{1} \right> } \right|$ there are at least a multiple of $p$ many tuples whose components are all equal, whether it be $(e, e, \cdots , e)$ or $(g , g , \cdots , g )$. But the very fact that such elements belong to $\left| X_{ \left< \rho_{1} \right> } \right|$ means that $g \cdots g = g^p = e$. Therefore, at the very least, we can confirm that $\left< g \right>$ satisfying $\left< g \right> = p$ is a subgroup of $G$.

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