Can we give of the fact that a group of order $9$ is abelian without using an argument involving the product of two cyclic groups of order $3$? A group of order $9$ is always abelian. I've seen proofs of this result, but I would like to prove it the following way: 
Let $G$ be a group of order $9$. If $G$ has an element $a$ of order $9$, then $G$ is cyclic and therefore abelian. 
So let's assume that $G$ has no element of order $9$. Let $a$ be an element of $G$ other than the identity $e$. Then, due to a corollary of the Lagrange's theorem, the order of $a$ must be a divisor of $9$ and so must be $3$. 
We can also show that the order of $a^2$ is also $3$. So we have the cyclic group $H_1 \colon= \langle a \rangle = \{e, a, a^2 \}$, which is a subgroup of $G$. 
Let $b$ be an element of $G$ that does not belong to $H_1$; of course $b$ exists since the order of $H_1$ is smaller than that of $G$, and $b$ is distinct from the identity $e$ since the latter is in $H_1$. 
As before, we can show that the order of $b$ is $3$, which is also the order of $b^2$, and that we have another cyclic subgroup of $G$ given by $H_2 \colon= \langle b \rangle = \{e, b, b^2 \}$. 
What is the intersection of $H_1$ and $H_2$? Can we show that $H_1 \cap H_2 = \{ e \}$? 
Now let's form the subset product $H_1 H_2$. 
$$ H_1 H_2 = \{ xy \ | \ x \in H_1, y \in H_2 \}. $$ So the number of elements in $H_1 H_2$ is $3 \times 3 = 9$. Right? Is this reasoning correct? 
Since the cardinality of this subset product set is the same as that of $G$, this product of subsets must be all of $G$. 
Hence $H_1 H_2$ is a subgroup of $G$. So we can show that $$H_1 H_2 = H_2 H_1.$$ 
Now we have to show that $ab = ba$, which is to be done by showing that $ba$ cannot be equal to any other element of $H_1 H_2$. Once we have done that, we would have effectively shown that $G$ is abelian. 
Is this reasoning correct? If so, can anyone please fill in the holes, if any? 
 A: This isn't exactly an answer to your question, but shows how to put this example into a more general context.
You can use two more general results, which prove that a group $G$ of order $p^2$ is always abelian when $p$ is a prime.


*

*The centre $Z$ of $G$ is non-trivial. Here one uses the fact that the sizes of the conjugacy classes in $G$ divide the order of the group and therefore have size $1$ or $p$. Since the conjugacy class of the identity has size $1$, if all the other classes has size $p$ counting the elements of the group would give $1+mp=p^2$ for some integer $m$. But this is impossible. A modification of this proof works for any group of order $p^n$ with $n\ge 1$.

*If $G/Z$ is cyclic, then $G$ is abelian. Here we take $aZ$ as a generator of the cyclic group, and show that any element of $G$ can be written as $a^rz$ for some integer $r$ and some $z\in Z$. Then $a^rz_1a^sz_2=a^sz_2a^rz_1$ using the fact that $z_1, z_2$ are in the centre.


Then it follows that a group $G$ of order $p^2$ has non-trivial centre, so $G/Z$ has order $1$ or $p$, so must be cyclic, so $G$ must be abelian.
I've given the shape of the argument and left some details for you to fill in.
