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
 A: You have a finite group $G$ and you take any element $g\in G$. Then $\langle g \rangle$ is a subgroup of $G$. Then, as mentioned in the comment by anon, you can apply Lagrange's theorem to get the conclusion that you want.
As an example of this, you could consider the symmetric group $S_5$. You pick a random element $\sigma \in S_5$, for example $\sigma = (1, 2, 4)$. Then you get the  subgroup
$$
\langle \sigma\rangle = \{(1,2,4), (1, 4, 2), (1) \}.
$$
Hence the order of $\langle \sigma\rangle$ is $3$, and indeed 3 is a divisor in  $O(S_5) = 5! = 120$.
You ask in the comment above about an example with a subgroup of the complex numbers. Consider $z = e^{\frac{2\pi i}{15}}$. Then you have the group $G = \langle z\rangle$ (under multiplication). This group has order $15$. Can you find/write down the elements?)Now take $w= e^{\frac{2\pi i}{5}}$. Then $\langle w\rangle$ is a subgroup of $G$ of order ... ( I will let you think about that).
As an application of this someone else might have something helpful to say. 
A: 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.
