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.

If you can show that $G$ has an element $g$ of order 3, then $\langle g \rangle$ is a normal subgroup (since $G$ is abelian), and therefore $G / \langle g\rangle$ is an abelian group. Since the order of $G / \langle g\rangle$ is 33, this group also contains an element, say $x + \langle g\rangle$, of order 3. If $n$ is the order of $x$ in $G$, then $(x + \langle g\rangle)^n = e$, so $n$ is divisible by 3. Therefore, $y = x^{n/3}$ is an element of $G$ with order $3$. Now you can check that the subgroup of $G$ generated by $g$ and $y$ has order 9.
So the problem reduces to showing that $G$ (or any finite group with order divisible by 3) has an element of order 3. This is proved quite similarly to the argument above, and essentially amounts to proving Cauchy's theorem for the special case of an abelian group. This is done adequately by Wikipedia, so I won't reproduce it here.
