Order of coset is greater than number of cosets in $\mathbb{Z}_2\times\mathbb{Z}_4/\langle(0,1)\rangle$

$$\mathbb{Z}_2\times\mathbb{Z}_4/\langle(0,1)\rangle$$

Since $\langle(0,1)\rangle$ has order $4$ and $\mathbb{Z}_2\times\mathbb{Z}_4$ has 8 elements, the quotient has $\frac{8}{4} = 2$ elements. However, when I calculate the order of the coset:

$$(0,1)+\langle(0,1)\rangle$$

I have the following elements:

$$(0,1)+\langle(0,1)\rangle$$ $$(0,2)+\langle(0,1)\rangle$$ $$(0,3)+\langle(0,1)\rangle$$ $$(0,0)+\langle(0,1)\rangle$$

so it must have order $4$

But the quotient is a group with just $2$ cosets. How is it possible?

Am I confusing something? I know that the group generated by some element must have an order that divides the order of the group, which is $2$. So why it has order $4$?

• You've listed the same element 4 times. You want to look at $(0,0)$ and $(1,0)$. – Ted Shifrin Apr 28 '17 at 0:55
• So how do I calculate the order of a coset? – Guerlando OCs Apr 28 '17 at 0:56
• I must sum things until they become another coset? – Guerlando OCs Apr 28 '17 at 0:59
• The only nonzero coset here is $(1,0)+\langle (0,1)\rangle$. – Ted Shifrin Apr 28 '17 at 1:01
• @TedShifrin Now suppose $(\mathbb{Z}_2\mathbb{Z}_4)/\langle(1,2)\rangle$, the elements in the $0$ coset are just $(0,0),(1,2)$. So if I take $(0,1)+\langle(1,2)\rangle$ and add itself to it more times, I get: $(0,2),(0,3),(0,0)$, so the order of this coset is $4$, right? – Paprika Apr 28 '17 at 1:11

$\langle (0,1) \rangle=\{(0,0),(0,1),(0,2),(0,3)\}$.
If $g\in H$, then $g+H=H$
So the first coset is $\langle (0,1) \rangle$. Since the subgroup has only two cosets, so choose an element not in $\langle (0,1) \rangle$ to form another coset. To be easy, we can choose $(1,0)$ so that $(1,0)+\langle (0,1) \rangle$ is another coset.