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I'm trying to find the order of this factor group: $$(\mathbb Z_{12}\times\mathbb Z_{18}) / \langle (4,3)\rangle.$$

The order of the factor group is just the number of elements in it (aka the number of cosets). Let $H = \langle(4,3)\rangle, G = Z_{12}\times Z_{18}$.

$$H = \{(4,3), (8,6), (0,9), (4,12), (8,15), (0,0)\}$$

So a coset of $H$ in $G$ is of the form $aH$, where $a \in G$

Let $a = (0,0)$. Then $aH = H$. From above, we know this coset has 6 elements. Further, each coset must have the same number of elements.

So now we just take the order of G and divide by 6 to get the order of $G$ / $H$.

There are 12 choices for $Z_{12}$ and 18 choices for $Z_{18}$ so order of $G = 12 \cdot 18 = 216$.

$216 \div 6 = 36$. So the order of $G / H$ is 36.

Does this look right?

Thanks for the help, Mariogs

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    $\begingroup$ Yep. All correct. :) $\endgroup$ – Klaas van Aarsen Mar 29 '14 at 20:42
  • $\begingroup$ @IlikeSerena hello, how can I get the elements of the factor group? $\endgroup$ – Salvattore Jun 6 '16 at 17:06
  • $\begingroup$ Hi @Salvattore. They are the elements $aH$ for the different values of $a$. That is (0,0)H, (1,0)H, (2,0)H, (3,0)H, etcetera. We can skip e.g. (4,0)H now, because that is the same as (0,0)H. $\endgroup$ – Klaas van Aarsen Jun 9 '16 at 17:55
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    $\begingroup$ Thanks @IlikeSerena, I will find all elements right now!!! hahahahaha $\endgroup$ – Salvattore Jun 9 '16 at 20:11
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    $\begingroup$ @Salvattore, I think that should be aHaHaHaHaH. :D $\endgroup$ – Klaas van Aarsen Jun 10 '16 at 21:28

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