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I know that the set of all equivalence classes Z/nZ is a group (with identity element the equivalence class [0], inverse element -[a]=[n-a]=[-a], etc.).

However, is a single equivalence class modulo n, say for n=5 and the class [3], a group? It contains the elements {...,-7,-2,3,8,13,..} so I don't see how it could possibly be a group since it doesn't contain 0 (if it is, then what are its identity and inverse elements)?

I think it's clear the answer is no. However, I raise this question because this video (http://www.youtube.com/watch?v=eooojREZMfo) around the 21 minute mark seems to suggest that the cosets of Z/nZ are subgroups of the additive integers (where the cosets a+nZ are the equivalence classes modulo n, namely [0]=nZ, [1]=1+nZ, [2]=2+nZ,..., all the way up to the class [n-1]).

Thanks for your responses!

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No closure under addition in the eq. class. $3+8=11$, if you are talking addition. $3\cdot{8}=24$ if you are talking multiplication –  Eleven-Eleven Oct 11 '13 at 2:20
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"seems to suggest the cosets of Z/nZ are subgroups [themselves]" I don't see how you got this impression; nothing the lecturer said suggests that in any way. –  anon Oct 11 '13 at 2:36

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The video states that "the set of all cosets forms a group". This is exactly the set $\mathbb{Z}/n\mathbb{Z}$ as you know it. It doesn't say that a coset itself is a subgroup of $\mathbb{Z}$. And indeed, the coset $3 + 5\mathbb{Z}$ is not a subgroup of $\mathbb{Z}$, as you noted. The set $3 + 5\mathbb{Z}$ could be turned into a group though, if we define some silly operation on it that has the identity element $-12$ for example.

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Thanks for the clarification! –  Mathemanic Oct 11 '13 at 2:49

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