I need help solving the following problem:
Let $G = \Bbb{Z}_9 \times \Bbb{Z}_9 \times \Bbb{Z}_9$ be the product of three cyclic groups of order 9. Give invariant factors and elementary divisors of the quotient group $G/H$ when
a) $H= \langle (6, 6, 6)\rangle \subset G$
b) $H = \langle (1, 3, 0), (1, 0, 3) \rangle \subset G$
I only got so far as to determine the order of the quotient group in each case: for part a), I know $H=|\langle (6,6,6)\rangle| =3$, so $G/H$ has order $729/3=243$ and at least intuitively, this seems to point in the direction of $G/H \cong \Bbb{Z}_9 \times \Bbb{Z}_9 \times \Bbb{Z}_3$, but I am not sure about this.
For part b), $|H|=|\langle (1, 3, 0), (1, 0, 3) \rangle|=9$, but I don't know how to proceed from here, as $|G/H| = 81$ in this case, but I don't know what to do next to determine the correct decomposition.
Is there any consistent method of solving problems like this one? Any help would be greatly appreciated.