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Have I done this question correctly?

I have worked out that the quotient group $\mathbb{Z}$12/$\mathbb{Z}$3 = $\mathbb{Z}$3. ie the group {(0,1,2)}.

For my cosets of $\mathbb{Z}$3 I had three of the same sets (all a variation of {0,1,2} ) and $\mathbb{Z}$12 is {0,1,5,7,11}

I can't help but feel that this doesn't look correct.

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    For starters, the order of the quotient group must be $|\mathbb{Z}_{12}/\mathbb{Z}_3|=|\mathbb{Z}_{12}|/|\mathbb{Z}_3|=12/3=4$... – 2017-01-09
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    Yes that's why I found it strange that I got (0,1,2). Could you offer some insight into where I went wrong? @kccu – 2017-01-09
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    What are the cosets? There should be 4 of them. And I don't understand what you mean when you say "$\mathbb{Z}_{12}$ is $\{0,1,5,7,11\}$." – 2017-01-09
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    the group $\mathbb{Z}_12$ under addition mod 12 is {0,1,5,7,11}, right? and since $\mathbb{Z}_3$ is {0,1,2} the cosets are {0+1,0+5,0+7...0+11},{1+0,1+1,1+5...1+11}, {2+0,2+1,...2+11} – 2017-01-09
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    The group $\mathbb{Z}_{12}$ is $\{0,1,2,3,4,5,6,7,8,9,10,11\}$. I'm not sure where your numbers came from. Cosets of $\mathbb{Z}_3$ are of the form $a+\mathbb{Z}_3$ for some $a \in \mathbb{Z}_{12}$. E.g., $0+\mathbb{Z}_3$, $1+\mathbb{Z}_3$, $\dots$, $11+\mathbb{Z}_3$. Some of these will be the same, but there will be 4 distinct ones, which are the elements of $\mathbb{Z}_{12}/\mathbb{Z}_3$. – 2017-01-09
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    Ah I see where I went wrong. But I still don't understand why I am only getting 3 distinct cosets. I have 0+$\mathbb{Z}_3$, 1+$\mathbb{Z}_3$, 2+$\mathbb{Z}_3$, and 3+$\mathbb{Z}_3$ but isn't this last one mod 3 so simply is equal to 0+$\mathbb{Z}_3$? – 2017-01-09
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    $\mathbb{Z}_3$ is not a subgroup of $\mathbb{Z}_{12}$ in the way you have it identified. You need to find a subgroup of $\mathbb{Z}_{12}$ isomorphic to $\mathbb{Z}_{3}$ – 2017-01-09
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    Right, as @PaulPlummer noted, you can only take the quotient by a subgroup of $\mathbb{Z}_{12}$. But $\{0,1,2\}$ is not a subgroup of $\mathbb{Z}_{12}$, so you need to identify $\mathbb{Z}_3$ with an isomorphic subgroup of $\mathbb{Z}_{12}$. (Hint: choose an element of $\mathbb{Z}_{12}$ of order $3$ to generate the subgroup.) – 2017-01-09

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For your information, $\mathbb Z_{12}$ is the integers $0,1,\ldots,11$ under addition mod 12, while $\mathbb Z_3$ is the integers $0,1,2$ under addition mod 3; the latter is not a subgroup of the former.

Given $\mathbb Z_{12}$, are you thinking of the subgroup $\langle3\rangle=\{0,3,6,9\}$? In this case, note that it is a subgroup of order 4. On the other hand, if you're thinking of a subgroup of order 3, then it would be $\langle4\rangle=\{0,4,8\}$.

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    I'm still a bit confused...how can 4 be a generator if we are just considering $\mathbb{Z}_3$ (where $\mathbb{Z}_3$ is (0,1,2) )? – 2017-01-09
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    As I said, $\mathbb Z_3$ is _not_ a subgroup of $\mathbb Z_{12}$; $\mathbb Z_{12}/\mathbb Z_3$ does not make sense. My guess is that you are thinking of either $\mathbb Z_{12}/\langle3\rangle$ or $\mathbb Z_{12}/\langle4\rangle$. – 2017-01-09
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    Okay thanks a lot, I guess it is a badly written question then. – 2017-01-09