Group Z/3Z Proof of Subgroup.

Question

I am looking at the quotient group G = Z/3Z. 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. 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?