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

Proof of Cauchy's Theorem in group Theory

Theorem ¹

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

Explanation

When people usually refer to Cauchy’s theorem, they don’t typically mean this theorem. Another Cauchy’s theorem, which is fundamental to complex analysis, is much more significant and often mentioned. More so because this theorem is generally superseded by the First Sylow theorem, making it quite rare to specifically use Cauchy’s theorem.

While it might not necessarily be useful to know, the method of proof is very unique, starting from its setup. It is recommended to try to prove it at least once, even if just for interest.

Proof

Consider $i=1 , \cdots p$ where $g_{i} \in G$ 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 $S_{p}$.

$\rho_{1} \in S_{p}$ acts as a permuting permutation that shifts each tuple of $X$ by one place, $$ \rho_{1} (g_{1}, g_{2} , \cdots , g_{p-1} , g_{p}) = (g_{2}, g_{3} , \cdots , g_{p} , g_{1}) $$ in such action. Defined as $g_{p} = (g_{1} \cdots g_{p-1} )^{-1}$, therefore $|X| = |G|^{p-1}$ and since $p$ is a divisor of $|G|$, $p$ is also a divisor of $|X|$.

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

Hence $| \left< \rho_{1} \right> | = p$, $$ |X| \equiv \left| X_{ \left< \rho_{1} \right> } \right| \pmod{p} $$ and $p$ also divides $\left| X_{ \left< \rho_{1} \right> } \right|$. This means that there are at least as many tuples in $\left| X_{ \left< \rho_{1} \right> } \right|$ with all elements being the same, regardless of whether it becomes $(e, e, \cdots , e)$ or $(g , g , \cdots , g )$, as multiples of $p$. However, the very presence of these elements in $\left| X_{ \left< \rho_{1} \right> } \right|$ signifies $g \cdots g = g^p = e$. Therefore, one can confirm that at least $\left< g \right> = p$ satisfying $\left< g \right>$ will be a subgroup of $G$.

■