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.
Best Answer
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?