Find the order of $(3,1)+\langle0,2\rangle$ in $ \Bbb Z_{4}\oplus \Bbb Z_{8}/\langle0,2\rangle$ QuestionLet $\langle 0,2\rangle $ denote the sugroup of $\Bbb Z_{4}\oplus \Bbb Z_{8}$ generated by $(0,2)$. Then find the order of $(3,1)+\langle0,2\rangle$ in $\Bbb Z_{4}\oplus \Bbb Z_{8}/\langle0,2\rangle$
MY Approach $|(0,2)|$ in $\Bbb Z_{4}\oplus \Bbb Z_{8}$ is 4
$$|\Bbb Z_{4}\oplus \Bbb Z_{8}/\langle0,2\rangle|= 8\\
(3,1)+\langle0,2\rangle=\{(3,3),(3,5),(3,7),(3,1)\} $$
i don't know the formula to calculate the order of this coset
 A: Well, we have
$$
\big((3,1)+\langle0,2\rangle\big) + \big((3,1)+\langle0,2\rangle\big) = (2,2)+\langle0,2\rangle\\
=\{(2,2),(2,4),(2,6),(2,0)\}\neq (0,0) + \langle 0,2\rangle
$$
so the order is not $2$. Now check whether the order is $3, 4$ and so on. It won't take long until you have your answer.

This can be made somewhat shorter if you note that the order of $(3,1)+\langle0,2\rangle$ must be a divisor of the order of $(3, 1)\in \Bbb Z_4\oplus\Bbb Z_8$, which is $8$. So the answer cannot be $3$, and you can skip that and go straight to $4$. If the order isn't $4$, then it can't be $5, 6$ or $7$, so it must be $8$, and you're done.

A: $1 \cdot (3,1) = (3,1) \notin \langle 0,2\rangle$
$2 \cdot (3,1) = (2,2) \notin \langle 0,2\rangle$
$3 \cdot (3,1) = (1,3) \notin \langle 0,2\rangle$
$4 \cdot (3,1) = (0,4) \in \langle 0,2\rangle$
Hence, the order of $(3,1)+\langle0,2\rangle$ is $4$.
A: $|(3,1)+\langle (0,2)\rangle|\in \{1,2,4,8\}$. 
Here $|(3,1)+\langle (0,2)\rangle|\neq 1$ as $(3,1)+\langle (0,2)\rangle \neq \langle (0,2)\rangle$ and $2[(3,1)+\langle (0,2)\rangle]=(6,2)+ \langle (0,2)\rangle=(0,0)+\langle (0,2)\rangle$. 
Hence the required answer is 2.
