How to prove something about the order of element at a group $G$ is a cyclic group. the order of $G$ is $n$ ($|G|=n$).
$m\mid n$,
I have to prove that there is $b\in G$ that $ord(b)=m$.
Know, here is one part of the proof:
if $m\mid n$ that means that $n=mk$. Lets mark:
$$b=a^k$$ 
then: $$b^m=(a^k)^m=a^n=e\Rightarrow ord(b)=m$$
Now I need to prove that $m$ is minimal.
How I'm prove this?
Thank you!
 A: Answer to Original Post, Before Edit:
The theorem you are trying to prove, as stated, is not true. The converse of Lagrange's Theorem does not, in general, hold. 
For example, take $A_4$, the alternating group of $S_4$. $\;|A_4| = 12$, and clearly, $\;6\mid 12.\;$ However, there is no subgroup of order $6$ in $A_4$, and hence, no element $b\in A_4$ whose order is $6$. 
So Lagrange holds here, and the corresponding Corollary that for every element $b\in A_4$, the order of $b$ divides the order of $A_4$ is of course true, but it does not follow that if $m$ divides the order of a group, then there is an element of order $m$.
With limitations, the theorem is true. Specifically, 


*

*if $G$ is a finite abelian group, or

*if the order of $G$ is a power of a prime, or 

*if $G$ is solvable,
then the statement is true.

After edit: Indeed, this is true for the finite cyclic group $G$.  
Suggestion about the initial part of your proof: Make sure you make explicit that, since G is cyclic, it is generated by one element, we'll call $a$. So $\langle a\rangle = G$. Then it follows that there is some element $b \in G$ such that $b = a^k$, where $0\leq k \lt n$.
To show that $m$ is minimal, we can show that if there exists $0\leq s \lt m$ such that $b^s = e$, then $s = 0$. Or else you can assume, for the sake of contradiction, that there exists an $s$, $0\lt s \lt m$ such that $b^s = e$, and obtain a contradiction.
