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

• nigelscott
In summary, the conversation discusses the proof of [0] being a normal subgroup in the quotient group G = Z/3Z, which is additive and abelian. The group operation is written as multiplication and the identity element 0 is also in G, leading to the conclusion that ene-1 = n applies to multiplicative groups where e = 1. The convention is to write a group multiplicative, but in this case, it is also possible to use equivalence classes [0],[1],[2] to represent the elements.
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?

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.

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.

nigelscott said:
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\,\},+)##.

Thanks, makes more sense now.

What is a subgroup?

A subgroup is a subset of a group that satisfies the same group axioms as the original group. This means that it is closed under the group operation, contains the identity element, and every element has an inverse within the subgroup.

What is Z/3Z?

Z/3Z refers to the group of integers modulo 3, also known as the cyclic group of order 3. This group has three elements: {0, 1, 2} and follows the group operation of addition modulo 3.

How do you prove that a subset is a subgroup of Z/3Z?

To prove that a subset is a subgroup of Z/3Z, you need to show that it satisfies the subgroup criteria of being closed under the group operation, containing the identity element, and having inverses for all elements. Additionally, you can use the subgroup test, which states that if the subset is non-empty and the product of any two elements in the subset is also in the subset, then it is a subgroup.

What is the identity element in Z/3Z?

The identity element in Z/3Z is 0, since any element added to 0 results in that same element. This follows the definition of the identity element in a group, which is an element that, when combined with any other element, results in that same element.

Can there be more than one subgroup of Z/3Z?

Yes, there can be multiple subgroups of Z/3Z. Since a subgroup is a subset of a group, any subset that satisfies the subgroup criteria can be considered a subgroup. This means that there can be multiple subgroups of different sizes and elements within Z/3Z.

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