Show that $9 \mid |G|$ when $G$ has at least $3$ elements of order $3$, $G$ - finite abelian group

Suppose that a finite abelian group $G$ has at least three elements of order $3$. Prove that $9$ divides $|G|$.

My attempt -

Let $a$ and $b$ be the elements of $G$, such that $|a| = |b| = 3$. Then $$\{ 1,a,a^2,b,b^2,ab,ab^2,a^2b,a^2b^2 \}$$ is a subgroup of $G$. And, by applying Lagrange's theorem we get $9 \mid |G|$.

Doubt - What is the significance of having at least three elements of order $3$? In the group constructed above there are more than three elements of order $3$.

Any hints or suggestions will be appreciated.

• You need to assume that $b \ne a$ and $b \ne a^2$, and you can choose such a $b$ if there are at least three elements of order $3$. – Derek Holt Jun 17 '18 at 12:51
• How do you know that all the elements you have written down in the subgroup are distinct? You will need to use the hypothesis in proving that the elements are distinct. – Brahadeesh Jun 17 '18 at 12:58
• Ok thanks ..... – blue boy Jun 17 '18 at 14:43
• In fact, any group has to have an even number of elements of order $3$. This is because $a^2=a^{-1}$ has order $3$ when $a$ does... See: math.stackexchange.com/questions/256631/… – Chris Custer Jun 17 '18 at 16:24

Your argument is incomplete. But the statement says “at least $3$”, so when you find $3$ of them you're done and the presence of further elements is irrelevant.
A finite abelian group is a direct sum of cyclic groups of prime power order. Since $3$ divides the order of the group, there must be at least a summand of the form $\mathbb{Z}/3^k\mathbb{Z}$. If $k\ge2$ we are done.
Otherwise, only summands of the form $\mathbb{Z}/3\mathbb{Z}$ appear; if there are two of them, we're done. So the case to be examined is $$G=\mathbb{Z}/3\mathbb{Z}\oplus H$$ where $3$ does not divide $|H|$. How many elements of order $3$ does this group have?