Proving Normality of [0] in Z/3Z Quotient Group

[nigelscott]

Homework Statement

I am looking at the quotient group G = Z/3Z which is additive and abelian. The equivalence classes are:

[0] = {...,0,3,6,...}

[1] = {...,1,4,7,...}

[2] = {...,2,5,8,...}

I want to prove [0] is a normal subgroup, N, by showing gng^-1 = n' ∈ N for g ∈ G and n ∈ N. Since G is abelian so gg^-1n = n' ∈ N. The identity element 0 is also in G so I should be able to write 0.0^-1n = n'. How do I interpret 0.0-1?

Homework Equations

The Attempt at a Solution

My first thought was that since the inverse of the identity is the identity then 0.0^-1 = 0. Therefore, this would give 0n = n' = 0. This is also consistent with 0n = n0. However, I am still not sure about this because the identity is being used multiplicatively and not additively.

 

[fresh_42]

You should not use $0$ if you write the group operation as multiplication, in which case the elements are $\{\,1,2,3\,\}$. If you write it as addition, then the normality condition will be $x+N-x\ \subseteq N$.

 

[nigelscott]

OK. So is it fair to generalize by saying that ene^-1 = n applies to multiplicative groups where e = 1 and that e + n + (-e) = n applies for additive groups where e = 0? This makes intuitive sense but most of the literature I have read seems to focus on the multiplicative case.

 

[fresh_42]

OK. So is it fair to generalize by saying that ene^-1 = n applies to multiplicative groups where e = 1 and that e + n + (-e) = n applies for additive groups where e = 0? This makes intuitive sense but most of the literature I have read seems to focus on the multiplicative case.

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: $[0],[1],[2]$ and 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 $\mathbb{Z}_3 = (\{\,0,1,2\,\},+)$.

 

[nigelscott]

Thanks, makes more sense now.