Consider $G=\mathbb Z/12\times \mathbb Z/12$ and its subgroup $H$ generated by $(a^4,a^6)$, where $a$ is a generator of $\mathbb Z/12$. How do I find $G/H$ as a product of cyclic groups of prime power orders?
I know how to identify similar quotients in the case $\mathbb Z\times \mathbb Z$ via reducing a matrix to the Smith normal form (see this answer for instance), but I don't know whether such technique carries over to this case.
Note that $H=\langle(a^4,0),(0,a^6)\rangle$ (this is not trivial, so it has to be proven, but it's quite easy), and by the third isomorphism theorem, you can divide out by one element at a time. So we get $$\Bbb Z/12\times\Bbb Z/12/\langle(a^4,0),(0,a^6)\rangle\\\cong \big(\Bbb Z/12\times\Bbb Z12/\langle(a^4,0)\rangle\big)/\langle(0,a^6)\rangle\\\cong\Bbb Z/4\times\Bbb Z/6$$