Let $G = \mathbb{Z}_{12} \times \mathbb{Z}_{12}$, and let $a$ be a generator of $\mathbb{Z}_{12}$.
Consider the subgroup $H$ generated by $(a^4, a^6)$. I need to write $G/H$ as a product of cyclic groups each of which has order equal to a power of some prime.
My work
Here $|H| =6$, so by Lagrange's theorem we have $|G/H|=\dfrac{144}{6}=24=8\times 3 = 2^3 \times 3$.
Can I say that $G/H \cong \mathbb{Z}_8 \times \mathbb{Z}_3$?
By this standard result, a presentation for $G$ is
$$\langle x,y\mid x^{12}, y^{12}, xy=yx\rangle.$$
The element that generates $H$ corresponds to $x^4y^6$, so that, by definition of a presentation, we have
$$G/H\cong\langle x,y\mid x^{12}, y^{12}, xy=yx, x^4y^6\rangle.$$
Now we can perform Tietze transformations:
$$\begin{align} G/H&\cong\langle x,y\mid x^{12}, y^{12}, xy=yx, x^4=y^{-6}=y^6\rangle\\ &\cong \langle x,y\mid x^{12}, (y^6)^2, xy=yx, x^4=y^6\rangle\\ &\cong\langle x,y\mid x^{12}, (x^4)^2, xy=yx, x^4=y^6\rangle\\ &\cong\langle x,y\mid x^{\gcd(12,8)}=x^4=e, y^6=x^4, xy=yx\rangle\\ &\cong\langle x,y\mid x^4, y^6, xy=yx\rangle\\ &\cong\Bbb Z_4\times\Bbb Z_6\\ &\cong \Bbb Z_{12}\times\Bbb Z_2, \end{align}$$
where the final isomorphism is because $\Bbb Z_{12}\cong\Bbb Z_4\times\Bbb Z_3$ while $\Bbb Z_6\cong\Bbb Z_3\times \Bbb Z_2$.