How can $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)> \cong \mathbb{Z}_{12} = \mathbb{Z}_{4} \times \mathbb{Z}_{3}$?
I am not convinced at the least that $\mathbb{Z}_{12}$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$
For instance, doesn't $<1>$ have an order of 12 in $\mathbb{Z}_{12}$? And no element of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$ can even have an order of $12$ no?
What is the maximum order of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$? I know if the denominator isn't there, it is $lcm(4,6) = 12$, but even if it weren't there, I don't see how either component can produce an element of order 12.
What I mean is that the first component is in $\mathbb{Z}_4$, so all elements have max order 6 and likewise $\mathbb{Z}_6$, have order 6, so how can any element exceed the order of their group?