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Consider the following exercise:

Let $H \subset \mathbb{Z}^2$ be a subgroup generated by the vectors $2e_1$ and $2e_2$ (here $e_1$ and $e_2$ are generators of $\mathbb{Z}^2$). Consider the cosets: $\begin{pmatrix} 1 \\ 6\end{pmatrix} + H$, $\begin{pmatrix} 3 \\ 5\end{pmatrix} + H$ and $\begin{pmatrix} 7 \\ 11\end{pmatrix} + H$. How many of them are distinct in $\mathbb{Z}^2/H$?

I do not really understand what is asked. As far as I see it $H = \begin{pmatrix} 2n \\ 2k\end{pmatrix}$ with $n, k \in \mathbb{Z}$. Then certainly $\begin{pmatrix} 3 \\ 5\end{pmatrix} + H$ and $\begin{pmatrix} 7 \\ 11\end{pmatrix} + H$ would generate the exact same subset of $\mathbb{Z}^2$. For $\begin{pmatrix} 1 \\ 6\end{pmatrix} + H$ this is impossible however, since 6 is an even number and thus cannot become odd when adding multiplies of two to it (while adding multiplies of two to an odd number retains the oddness). So there would be two distinct elements ($ \in \mathbb{Z}^2/H$).

But this is extremely trivial, while the surround bulk of exercises is --- be it not difficult --- certainly not this easy. It confusing me. I would therefore really appreciate if someone could help me.

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    $\begingroup$ Your answer is fine. $\endgroup$ – Ravi Apr 10 '16 at 12:26
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Proof by mathematical democracy: I did this question yesterday and got exactly the same answer. :)

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