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.

There exists an element $a$ of order $3$. If there was not every element would need to have order $11$. 
But then a counting argument leads to contradiction.
We take the quotient group $ G/<a>$. Then it has order $33$. The same argument shows that there exists an element $b$ of order $3$. Take the map $p: G \rightarrow G /<a>$ the natural projection. The pre-image of a subgroup is a subgroup, so we take the subgroup of $G$ $p^{-1}(<b>)$. And this subgroup has order $9=|\rm{Ker p}|| <b>| $.  
