What is an intuitive way to think of Cauchy's theorem? I am looking at a problem which involves an understanding of why a finite group $G$ has an element with order $p$ if $p$ is a prime factor of $|G|$. I have looked at several resources and proofs online, and I only seem to find proofs and explanations which involve either strong induction or the orbit-stabilizer theorem. I find that the inductive proof doesn't give me much intuition of the theorem, and we haven't covered the orbit-stabilizer theorem in my class yet and in fact I'm only in week 2 of my group theory class. So I imagine there must be another simpler way to look at it.
I would be grateful for any alternative explanation of why there is such a relation between the factorization of a group's order and the order of its elements.
 A: There is a combinatorial proof that is quite easy to follow (and remember, too).
Assume that $G$ is a finite group and $p\mid |G|$. Let $X$ be the following set:
$$ X = \{ (g_1,g_2,\ldots,g_p) : g_1 g_2\cdot\ldots\cdot g_p = e\}.$$
We have $|X| = |G|^{p-1}$, since for every element of $X$ we are free to take the first $(p-1)$ coordinates as we want, then the last one is forced. So we have $p\mid |X|$.
Now we equip $X$ with an equivalence relation: we say that $x_1,x_2\in X$ are equivalent if there is some shift of the coordinates that brings $x_1$ into $x_2$. (Check it is an equivalence relation for real).
If the coordinates of $x\in X$ are not equal, the equivalence class of $x$ has exactly $p$ elements. On the other hand, if $x=(g,g,\ldots,g)$, then the equivalence class of $x$ is trivial (made of $x$ only) and $g^p=e$.
At last, we notice that the equivalence class of $(e,e,\ldots,e)$ is trivial. Since the equivalence classes form a partition of $X$ and $p\mid |X|$, there must be at least other $(p-1)$ elements of $X$ with a trivial equivalence class, so there are at least $(p-1)$ different elements of $G$ such that $g^p=e$ and $g\neq e$.
