Cauchy's theorem states that if $G$ is a finite group and $p$ is a prime dividing $|G|$ the order of $G$, then $G$ has a subgroup of order $p$. Such a subgroup is necessarily cyclic (proof).

Proof

The proof involves basically a single magic idea: from thin air, we pluck the definition of the following set.

Let $$X=\left\{\right(x_1,x_2,\dots ,x_p):x_1{x}_2\dots x_p=e\}$$ the collection of $p$-tuples of elements of the group such that the group operation applied to the tuple yields the identity. Observe that $X$ is not empty, because it contains the tuple $(e,e,\dots ,e)$.

Now, the cyclic group $C_p$ of order $p$ acts on $X$ as follows: $$(h,(x_1,\dots ,x_p\left)\right)\mapsto (x_2,x_3,\dots ,x_p,x_1)$$ where $h$ is the generator of $C_p$. So a general element $h^i$ acts on $X$ by sending $(x_1,\dots ,x_p)$ to $(x_{i+1},x_{i+2},\dots ,x_p,x_1,\dots ,x_i)$.

This is indeed a group action (exercise).

Now, fix $\overline{x}=(x_1,\dots ,x_p)\in X$.

By the Orbit-Stabiliser theorem, the orbit ${Orb}_{C_p}\left(\overline{x}\right)$ of $\overline{x}$ divides $|C_p|=p$, so (since $p$ is prime) it is either $1$ or $p$ for every $\overline{x}\in X$.

Now, what is the size of the set $X$?

Also, the orbits of $C_p$ acting on $X$ partition $X$ (proof). Since $p$ divides $|G|$, we must have $p$ dividing $|G{|}^{p-1}=|X|$. Therefore since $|{Orb}_{C_p}\left(\right(e,e,\dots ,e\left)\right)|=1$, there must be at least $p-1$ other orbits of size $1$, because each orbit has size $p$ or $1$: if we had fewer than $p-1$ other orbits of size $1$, then there would be at least $1$ but strictly fewer than $p$ orbits of size $1$, and all the remaining orbits would have to be of size $p$, contradicting that $p\mid |X|$.

Hence there is indeed another orbit of size $1$; say it is the singleton $\left\{\overline{x}\right\}$ where $\overline{x}=(x_1,\dots ,x_p)$.

Now $C_p$ acts by cycling $\overline{x}$ round, and we know that doing so does not change $\overline{x}$, so it must be the case that all the $x_i$ are equal; hence $(x,x,\dots ,x)\in X$ and so $x^p=e$ by definition of $X$.