Cauchy's theorem is the first major existence theorem in finite group theory. The focus keyword is Cauchy's theorem, because the theorem says that whenever a prime $p$ divides the order of a finite group $G$, the group must contain an element of order $p$. Lagrange's theorem tells us that the order of a subgroup divides $|G|$; Cauchy's theorem gives a partial converse for prime divisors. This result begins the route toward p-groups, p-subgroups, and the Sylow theorems, where prime-power divisors control the internal structure of finite groups. :::definition[Element of Prime Order] Let $G$ be a group and let $p$ be a prime number. An element $a\in G$ is said to have prime order $p$ if $a^p=e$ and $a^k\neq e$ for every integer $k$ satisfying $1\leq k<p$. ::: :::definition[Subgroup Generated by an Element] Let $G$ be a group and let $a\in G$. The subgroup generated by $a$ is $$\begin{equation*} \langle a\rangle=\{a^n:n\in\mathbb{Z}\}. \end{equation*}$$ If $o(a)=p$, where $p$ is prime, then $\langle a\rangle$ is a subgroup of $G$ of order $p$. ::: :::theorem[Cauchy's Theorem] Let $G$ be a finite group and let $p$ be a prime number. If $p$ divides $|G|$, then there exists $a\in G$ such that $o(a)=p$. ::: :::proof Given that $G$ is a finite group and $p$ is a prime number such that $p$ divides $|G|$. To prove that there exists $a\in G$ such that $o(a)=p$. Let $$\begin{equation*} X=\{(x_1,x_2,\ldots,x_p)\in G^p:x_1x_2\cdots x_p=e\}. \end{equation*}$$ The first $p-1$ entries may be chosen arbitrarily, and then the last entry is determined by $$\begin{equation*} x_p=(x_1x_2\cdots x_{p-1})^{-1}. \end{equation*}$$ Therefore $$\begin{equation*} |X|=|G|^{p-1}. \end{equation*}$$ Since $p$ divides $|G|$, it follows that $p$ divides $|X|$. Let the cyclic group $C_p$ act on $X$ by cyclic permutation: $$\begin{equation*} (x_1,x_2,\ldots,x_p)\mapsto(x_2,x_3,\ldots,x_p,x_1). \end{equation*}$$ This operation preserves $X$, because if $x_1x_2\cdots x_p=e$, then $$\begin{align*} x_2x_3\cdots x_px_1&=x_1^{-1}(x_1x_2\cdots x_p)x_1\\ &=x_1^{-1}ex_1\\ &=e. \end{align*}$$ Every orbit under this action has size $1$ or $p$. An orbit has size $1$ exactly when $$\begin{equation*} (x_1,x_2,\ldots,x_p)=(a,a,\ldots,a) \end{equation*}$$ for some $a\in G$. Such a tuple lies in $X$ exactly when $a^p=e$. The tuple $(e,e,\ldots,e)$ is one fixed point. Since $|X|$ is divisible by $p$ and every non-fixed orbit has size $p$, the number of fixed points is divisible by $p$. Therefore there is at least one fixed point different from $(e,e,\ldots,e)$. Thus there exists $a\neq e$ such that $a^p=e$. Since $p$ is prime, the order of $a$ is $p$. Hence there exists $a\in G$ such that $o(a)=p$. $\square$ ::: :::corollary Let $G$ be a finite group and let $p$ be a prime number. If $p$ divides $|G|$, then $G$ has a subgroup of order $p$. ::: :::proof Given that $G$ is a finite group, $p$ is a prime number, and $p$ divides $|G|$. To prove that $G$ has a subgroup of order $p$. By Cauchy's theorem, there exists $a\in G$ such that $o(a)=p$. Therefore $$\begin{equation*} |\langle a\rangle|=o(a)=p. \end{equation*}$$ Hence $\langle a\rangle$ is a subgroup of $G$ of order $p$. $\square$ ::: :::example Let $G=S_3$. Since $|S_3|=6$ and $2$ divides $6$, Cauchy's theorem guarantees an element of order $2$. The transposition $(12)$ satisfies $(12)^2=e$, so $o((12))=2$. Since $3$ divides $6$, Cauchy's theorem also guarantees an element of order $3$. The cycle $(123)$ satisfies $(123)^3=e$ and $(123)\neq e$, so $o((123))=3$. Hence Cauchy's theorem is verified in $S_3$ for both prime divisors of $|S_3|$. ::: This preview turns Cauchy's theorem into a prime-divisor checklist. Enter the order of a finite group and the visual marks every prime divisor of that order. Each highlighted prime is an element order that Cauchy's theorem guarantees somewhere inside the group. Try the presets $6$, $45$, $84$, and $77$ to compare the examples and exercises. Notice that the theorem guarantees prime orders, not composite orders such as $4$. :::scientific-preview[Cauchy Prime Divisor Explorer] ::: Each prime displayed is a forced element order, and the cyclic subgroup generated by that element has the same prime order. The theorem is deliberately limited to prime divisors, so larger divisor questions require stronger tools. This calculator performs the arithmetic check behind Cauchy's theorem. Enter a group order and, optionally, a divisor you are curious about. The calculator lists the prime orders that must occur and tells you whether your divisor is covered by Cauchy's theorem. Use it to see why order $4$ is not guaranteed merely because $4$ divides a group order. :::calculator[Cauchy Guarantee Calculator]