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


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.


$1 \cdot (3,1) = (3,1) \notin \langle 0,2\rangle$

$2 \cdot (3,1) = (2,2) \notin \langle 0,2\rangle$

$3 \cdot (3,1) = (1,3) \notin \langle 0,2\rangle$

$4 \cdot (3,1) = (0,4) \in \langle 0,2\rangle$

Hence, the order of $(3,1)+\langle0,2\rangle$ is $4$.


$|(3,1)+\langle (0,2)\rangle|\in \{1,2,4,8\}$.

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.

  • $\begingroup$ Your answer contains latex error, please correct this, so I can read your answer properly $\endgroup$ – Mohan Sharma Feb 2 '18 at 8:45
  • $\begingroup$ This answer is about $(0,3)$ and not $(3, 1)$ which was in the original question. Also, $2[(0,3) + \langle 0,2\rangle] = (0,6)+\langle 0,2\rangle = \langle0,2\rangle$, so the order of $(0,3)+\langle0,2\rangle$ is actually $2$. $\endgroup$ – Arthur Feb 2 '18 at 8:59
  • $\begingroup$ Apologize to both of you. The edit has been made. $\endgroup$ – Anjan3 Feb 2 '18 at 9:31

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