TITLE: Order of group element divides order of finite group
QUESTION [8 upvotes]: Proving this can be done as follows: consider a finite group G and elements $g_i \in G$ for some integer $i$. Now consider $\langle g_i \rangle = \{g_i^n: n\geq 0\}$, a generator. It can be proved that $\langle g_i \rangle \leq G$ and that the order of $g_i$ is equal to the order of $\langle g_i \rangle$, so $|\langle g_i \rangle| \leq |G|$.
We can now use Lagrange's theorem which states that if $H \leq G$ then $|H|$ divides $|G|$ and we're done. But...
Is there a simpler way to prove this fact?

REPLY [2 votes]: Fix an index $i \in \mathbb{N}$. Let $\langle g_{i}\rangle$ act on $G$ by left multiplication. Notice that the orbits under the action of $\langle g_{i}\rangle$ partition the group $G$. Let their number equal $k$. Now, notice that all orbits have cardinality $|\langle g_{i}\rangle|$ (Prove that for every $x \in G$ the map $\langle g_{i}\rangle \rightarrow \mathcal{O}(x)$ ,where $\mathcal{O}(x)$ is the orbit of $G$ that contains $x$, defined by $g_{i}^{n} \mapsto g_{i}^{n}x$ is a bijection). So,
$$|G|=\sum_{j=1}^{k}|\mathcal{O}_{j}|=\sum_{j=1}^{k}|\langle g_{i}\rangle|=k|\langle g_{i}\rangle|$$
and as a result the order of $g_{i}$ divides the order of the group $G$.
(It is practically the same answer as $\textbf{mesel's}$ but more detailed. I hope that it helped you.)