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How many cosets does the quotient group $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$ have

By definition, a coset is a set composed of all additions obtained by adding each element of a subgroup in turn by one particular element of the group, $G$, containing the subgroup, $H$. We can see, below, that there are $n$ left cosets. \begin{equation*} \begin{aligned} 0+n\mathbb{Z} & =\{0+nk | k\in \mathbb{Z}\} \\ 1+n\mathbb{Z} & =\{1+nk | k\in \mathbb{Z}\} \\ \vdots & = \vdots \\ n-2+n\mathbb{Z} & =\{n-2+nk | k\in \mathbb{Z}\} \\ n-1+n\mathbb{Z} & =\{n-1+nk | k\in \mathbb{Z}\} \\ \end{aligned} \end{equation*} Similarly, there are $n$ right cosets. However since addition is commutative the left and right cosets are equivalent. This implies that \textbf{there are $n$ cosets in the quotient group}.

Is there a more formal way to show that I have stated above?

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    $\begingroup$ Maybe use the division algorithm? cosets correspond to all possible remainders when you divide by $n$. $\endgroup$ – CPM Sep 8 '15 at 2:13
  • $\begingroup$ Perhaps contradiction is a good tool here. Suppose you have another representative not already in your list. $\endgroup$ – James S. Cook Sep 8 '15 at 2:18
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    $\begingroup$ I have edited the question to add the proof-writing tag. $\endgroup$ – user122283 Sep 8 '15 at 2:24
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As suggested in the comments, you could show that for any $k \in \Bbb Z$ that:

$k \in r + n\Bbb Z \iff k = qn + r$ with $r \in \{0,1,\dots,n-1\}$.

The advantage to this is that $r$, for any $k$, is unique, so that your list is thus complete (there are no more cosets than what you have listed), and that each coset is distinct (and necessarily disjoint from the others).

As it stands, you have just listed $n$ sets. Of course, you may have proven other theorems before that show that cosets are necessarily disjoint, but you still have to show you've listed ALL of them (that is, that every integer lies in one of the cosets you've listed).

I don't mean to sound overly harsh, because you've got the right idea-there are indeed $n$ cosets, which are precisely the sets you've listed.

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  • $\begingroup$ Do you think Induction is out of the question. Or I would be better off doing the proof by contradiction? $\endgroup$ – Username Unknown Sep 8 '15 at 13:31

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