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?

• You can certainly show that $H_1\cap H_2 = \{e\}$. If it isn't, it must contain either $b$ or $b^2$, as it is a subset of $H_1$. But then it must contain both, because an intersection of subgroups is always a subgroup. But then it is equal to $H_1$, so $H_2 = H_1$, which is a contradiction. – MJD Apr 19 '14 at 6:07
• @Saaqib Mahmuud We can also show that $H_1 H_2$ has $9$ elements by the following argument. Since $e,b,b^2$ are anyway distinct. Now consider $a^i b^j$=$a^k b^l$ such that all the powers are with respect to modulo 2. Now we can multiply by their inverses and obtain $a^x = b^y$ which is a contradiction until and unless $i=k and j=l$. But i am not sure how you would be able to show that $ab=ba$ – happymath Apr 19 '14 at 6:12
• happymath, you must be meaning modulo 3. Right? – Saaqib Mahmood Apr 19 '14 at 6:27
• @SaaqibMahmuud yeah i mean modulo 3 sry for the error – happymath Apr 19 '14 at 6:27

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.
1. 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 $n\ge 1$.
2. 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.