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I'm trying to find the factor group $\mathbb Z^2/H,$ where $H = \{(5k,3k):k\in\mathbb Z \}.$ Would the coset of $H$ containing $(a,b)$ simply be $\{(5k + a, 3k+b):k\in \mathbb Z\}?$ If so, then how could I generalize this to find the distinct cosets of $H?$

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  • $\begingroup$ Consider them separately at first. What would it mean to have $\mathbb{Z}/H$ where $H=\{5k : k\in \mathbb{Z}\}$? For a further hint, check out $\mathbb{Z}/5\mathbb{Z}$. $\endgroup$ – Eoin Oct 23 '14 at 2:53
  • $\begingroup$ Can you find a set of elements $\langle a,b\rangle$ of $\Bbb Z^2$ such that every element of $\Bbb Z^2$ is congruent to exactly one of them mod $H$? You already know, for instance, that such a set can’t contain both $\langle 1,1\rangle$ and $\langle 6,1\rangle$, since they’re congruent mod $H$ (i.e., differ by an element of $H$). $\endgroup$ – Brian M. Scott Oct 23 '14 at 2:55
  • $\begingroup$ @BrianM.Scott If I were to find such a set, that would give me the distinct cosets of $H.$ However, I'm not sure what you mean when you say that $(1,1)$ is congruent to $(6,1)$ mod $H.$ $(6,1) - (1,1) = (5,0) \notin H.$ $\endgroup$ – justin Oct 23 '14 at 3:34
  • $\begingroup$ Sorry: I misread the definition of $H$. Change $\langle 6,1\rangle$ to $\langle 6,4\rangle$. It may help to draw a picture: $H$ is a line through the origin in the integer lattice. Now think of the analogue from linear algebra: $H$ is like the vector subspace of $\Bbb R^2$ consisting of the line $y=\frac35x$. Cosets of $H$ are like lines parallel to that line. $\endgroup$ – Brian M. Scott Oct 23 '14 at 3:37
  • $\begingroup$ @BrianM.Scott Okay. then I think $\mathbb Z/H = \{H\} \cup \{(a,0) + H:a\neq 0\} \cup \{(0,b)+H:b\neq 0\}.$ Well, maybe not all of $\mathbb Z/H.$ But I think those are all distinct cosets. $\endgroup$ – justin Oct 23 '14 at 3:48
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Following the geometric suggestion in the comments above, $H$ is isomorphic to the intersection of $y=\tfrac{3}{5}x$ and the integer lattice. The distinct cosets of $H$ are then distinct lines parallel to this one (intersecting the integer lattice). Hence, $\mathbb Z^2/H = \{S_k:k\in \mathbb Z\},$ where $S_k = \{(a,b)\in \mathbb Z^2: b = \tfrac{3}{5}a + k\}.$

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  • $\begingroup$ Not quite, but you’re getting close. What about $\langle 1,1\rangle+H$? It’s the intersection with the integer lattice of the line $y-1=\frac35(x-1)$, or $5y-3x=2$, which doesn’t pass through any $\langle 0,b\rangle$ with $b\in\Bbb Z$. Note that all of the lines parallel to $H$ have equations of the form $5y-3x=k$. What’s the only restriction on $k$? $\endgroup$ – Brian M. Scott Oct 23 '14 at 4:08
  • $\begingroup$ @BrianM.Scott Oh! We need that $k\in \mathbb Z.$ Then $\mathbb Z^2/H = \{S_k:k\in \mathbb Z\},$ where $S_k = \{(a,b)\in \mathbb Z^2: b = \tfrac{3}{5}a + k\}.$ $\endgroup$ – justin Oct 23 '14 at 4:39
  • $\begingroup$ There you go. Once you fix that, you’ll have an answer that you can accept after the necessary time lapse for accepting one’s own answer. $\endgroup$ – Brian M. Scott Oct 23 '14 at 4:41

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