Find the largest possible order of an element of $\frac{\Bbb{Z}_{12}\times\Bbb{Z}_3\times\Bbb{Z}_6}{\left<(8,2,4)\right>}.$

Question

I find the following problem tough:

Find the largest possible order of an element of the quotient group

$$ Q = \frac{\mathbb{Z}_{12} \times \mathbb{Z}_{3} \times \mathbb{Z}_{6}}{\left<(8,2,4)\right>}. $$

So far I've got that the order of $\left<(8,2,4)\right>$ is $3$, so $|Q| = 72$. But I think I am lacking the basic intuition behind quotient groups involving direct product of groups. So the question has this second, added personlly, part: what is the intuition behind these quotient groups?

Thanks in advance!

Answer 1

$Q$ is the homomorphic image of $\Bbb Z_{12}\times\Bbb Z_3\times\Bbb Z_6$ under the canonical projection $p$. Any element of the image under a homomorphism must have order dividing $12$.

The only possibility for an element of order $12$ would be the image under $p$ of an element of order $12$. Consider $(1,0,0)$. It maps to $\overline{(1,0,0)}$, whose order is $12$.