$G$ be a group where $G= \mathbb{Z}_6 \oplus \mathbb{Z}_8$ and the normal subgroup $H=\langle(2,4)\rangle$ use orders of elements to determine...

Question

I have the elements of $H$: $\langle(2,4)\rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$ where $|G/H|=8$

possible isomorphic classes: $\mathbb{Z}_8$ ,$\mathbb{Z}_4\oplus \mathbb{Z}_2$, $\mathbb{Z}_2\oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$

I know

$\mathbb{Z}_6 \oplus \mathbb{Z}_8$ has an element of order $8$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $\mathbb{Z}_8$ or $\mathbb{Z}_4 \oplus \mathbb{Z}_2$.

I don't know how to find which one it is between those two. I know I need to compare the orders.

Answer 1

But $\mathbb Z_4\oplus\mathbb Z_2$ similarly has no element of order $8$.

The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.

So there's our answer: $\mathbb Z_4\oplus\mathbb Z_2$.