- #1

nigelscott

- 135

- 4

## 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

^{-1}n = n' ∈ N. The identity element 0 is also in G so I should be able to write 0.0

^{-1}n = 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<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.