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/&lt;a&gt;$. Then it has order $33$. The same argument shows that there exists an element $b$ of order $3$. Take the map $p: G \rightarrow G /&lt;a&gt;$ the natural projection. The pre-image of a subgroup is a subgroup, so we take the subgroup of $G$ $p^{-1}(&lt;b&gt;)$. And this subgroup has order $9=|\rm{Ker p}|| &lt;b&gt;| $.
If it's abelian then you know it's a product of cyclic groups.  You know that the order of these cyclic groups can be $3, 9, 11, 33$, or $99$ and their product must be $99$.  You've already seen that if $99$ is in their that you're done, and if $9$ is in their then you're done.  So your left with a product of cyclic subgroups of orders $3, 11$, or $33$ and their product must be $99$.

So show that $\mathbb Z/33$ has a subgroup of order $3$ and you must have either this group and $\mathbb Z/3$ or you must have two copies of $\mathbb Z/3$.  In either case this gives you a subgroup of order $9$.