# Proving Subgroup of Z/3Z

## Homework Statement

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

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

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

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

I want to prove  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?

## The Attempt at a Solution

My first thought was that since the inverse of the identity is the identity then 0.0<sup>-1</sup> = 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.

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fresh_42
Mentor
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 $x+N-x\ \subseteq N$.

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
Mentor
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: $,,$ and write the group operation as multiplication, but you will have $\cdot [x]=[x]$ and $^{-1}=$. 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 $,,$ 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 $\mathbb{Z}_3 = (\{\,0,1,2\,\},+)$.

Thanks, makes more sense now.