If $G$ is a finite group and $g \in G$, then $O(\langle g\rangle)$ is a divisor of $O(G)$ Does this result mean:


*

*Given any finite group, if we are able to find a cyclic group out of it (subgroup), then the order of the cyclic group will be a divisor of the original group.


If I am right in interpreting it, can one suggest an example of highlighting this? And also make me understand the possible practical uses of this result. It surely looks interesting
Thanks 
Soham

An application of this result is the formula
$$
\sum_{d\mid n} \phi(d) = n
$$
which can be estabilished by considering the cyclic group of order $n$: every element in this group has an order which is a divisor of $n$ and for every divisor $d$ of $n$ there are exactly $\phi(d)$ elements of order $d$.
A consequence of this formula is that finite multiplicative subgroups of a field are cyclic. In particular, the multiplicative group of a finite field is cyclic.
A simpler but very important consequence of the theorem is that groups of prime order are cyclic.
