Abelian group of order 99 has a subgroup of order 9 
Prove that an abelian group $G$ of order 99 has a subgroup of order 9.

I have to prove this, without using  Cauchy theorem. I know every basic fact about the order of a group.
I've distinguished two cases :


*

*if $G$ is cyclic, since $\mathbb Z/99\mathbb Z$ has an element of order $9$, the problem is solved.

*if $G$ isn't cyclic, every element of $G$ has order $1,3,9,11,33$. I guess I need to prove the existence of an element of order $9$. How should I do that ?
Note that $G$ is abelian (I haven't used it yet).
Context: This was asked at an undergraduate oral exam where advanced theorems (1 and 2) are not allowed.

Direct counting of elements can handle this without even using factor groups (although to understand group theory better, it's good to understand the answers here based on factor groups). 
There is one element of order $1$. This leaves $98$ elements to consider.
If even a single one is order $9$, it generates a subgroup of order $9$. 
If any one of them has order $99$, then the group is isomorphic to $C_{99}=\langle g\rangle$, and so $\{1,g^{11},\ldots,g^{88}\}$ is a subgroup of order $9$.
Otherwise, assuming no elements of order $9$ or $99$:


*

*Each element of order $33$ can be grouped with $\varphi(33)=20$ other elements of order $33$, taking powers relatively prime to $33$.

*Each element of order $11$ can be grouped with $\varphi(11)=10$ other elements of order $11$, taking powers relatively prime to $11$.


Taking elements of order $33$ and $11$ together, we would only have some multiple of $10$. Therefore at a minimum there are $8$ elements having order $3$. Take any two of these where neither equals the other squared, and since the group is abelian they generate a subgroup isomorphic to $C_3\times C_3$.
