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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.

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    $\begingroup$ You can use the Smith algorithm once you recognise you can step back and then consider $\mathbb{Z}\times\mathbb{Z}$ factored by $\langle (12,0), (0,12), (4,6)\rangle$. $\endgroup$ – ancientmathematician Jul 19 '18 at 6:46
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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 $$\Bbb Z/12\times\Bbb Z/12/\langle(a^4,0),(0,a^6)\rangle\\\cong \big(\Bbb Z/12\times\Bbb Z12/\langle(a^4,0)\rangle\big)/\langle(0,a^6)\rangle\\\cong\Bbb Z/4\times\Bbb Z/6$$

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