Which of the following are true about the group $\Bbb{Z}_6×\Bbb{Z}_9×\Bbb{Z}_{15}/\langle(5,5,3)\rangle$?

Question

Let, $\displaystyle{G=\frac{\Bbb{Z}_6\times\Bbb{Z}_9\times\Bbb{Z}_{15}}{\langle(5,5,3)\rangle}}$ Which of the following are ture about $G$? The given options are-
(a) $G$ is cyclic.
(b) $G$ is Abelian.
(c) $Z(G)$ is non-trivial.
(d) Order of $G$ is prime.
I have tried this-
Since $\Bbb{Z}_6\times\Bbb{Z}_9\times\Bbb{Z}_{15}$ is abelian, so the quotient group $G$ is abelian too. Hence, $G=Z(G)$. So (b), (c) are true.
Now, order of $\langle (5,5,3)\rangle$= the order of $(5,5,3)$=lcm$(6,9,5)=5\times9×2=90$. So order of $G=\frac{6×9×15}{90}=9=3^2$. Which is not prime. Hence, (d) is false.
But I cannot say anything about (a). A group of order $9=3^3$ may not be cyclic.
Can anyone prove or disprove option (a)? Thanks for assistance in advance.

Answer 1

$(2,0,0)+H,(0,3,0)+H$ and $(0,0,5)+H$ have order $3$ in the quotient. This implies $G$ is not cyclic.