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Proving Normality of [0] in Z/3Z Quotient Group
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[QUOTE="fresh_42, post: 6055775, member: 572553"] The convention is to write a group multiplicative, because they are usually not Abelian and we are used to associate commutativity with addition. So if a group is Abelian, then it is sometimes written as addition. But it's confusing to see things like ##x+N-x \subseteq N##. You can still use your notation of equivalence classes: ##[0],[1],[2]## [U]and[/U] write the group operation as multiplication, but you will have ##[0]\cdot [x]=[x]## and ##[2]^{-1}=[1]##. In this case it would be best to write ##\mathbb{Z}/3\mathbb{Z} = \mathbb{Z}_3 = \{\,1,a,a^2\,\}##. The reason is, that ##\mathbb{Z}_3## is even a ring and a field, so both operations are necessary. Of course the multiplicative group of ##\mathbb{Z}_3## as a ring or field doesn't have a zero in it, so we can use ##[0],[1],[2]## as elements. However, in your case, the group with three elements is considered, so either write ##\mathbb{Z}_3 = (\{\,1,a,a^2\,\},\cdot )## or ##\mathbb{Z}_3 = (\{\,0,1,2\,\},+)##. [/QUOTE]
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Proving Normality of [0] in Z/3Z Quotient Group
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