I know that the set of all equivalence classes $\Bbb Z/n\Bbb Z$ 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 $\Bbb Z/n\Bbb Z$ are subgroups of the additive integers (where the cosets $a+n\Bbb Z$ are the equivalence classes modulo $n$, namely $[0]=n\Bbb Z$, $[1]=1+n\Bbb Z$, $[2]=2+n\Bbb Z$,..., all the way up to the class $[n-1])$.
Thanks for your responses!
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.