Non-cyclic Finite abelian group generated by 2 elements

Question

I've thinking for a couple hours but I were not able to figure the following question:

Suppose I have a non-cyclic finite abelian group generated by two elements $a$ and $b$. Is it possible to have $(k,\ell)\notin \operatorname{ord} a\mathbb{Z}\times\operatorname{ord} b\mathbb{Z}$ such that $a^k b^\ell=e$?

Such $k$ and $\ell$ must satisfy $\operatorname{ord} a | bk$ and $\operatorname{ord} b | \ell a$.. and I also I thought about it with SmithNormalForm but I didn't get anywhere. Thanks in advance!

Answer 1

Yes. Consider $G_0=\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ with its “canonical basis” $(a_0,b_0)$ and $G=G_0/\langle 2,\, 2\rangle \cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $a,b$ are the images of $a_0,b_0$ in $G$. Then $2a+2b=0$ in $G$, but $a,b$ generate the noncyclic $G$ and both have order $4$.