my book states that if $N \trianglelefteq G$ then $(gN)^\alpha = g^\alpha N$ for $\alpha \in \mathbb{Z}$
The proof should be simple by induction but I can't understand how $(gN)^0 = [N = g^0N]$. How do you simplify $(gN)^0$? In my understanding that would be $\{(gn)^0 \ | \ \forall n \in N\} = \{1\} \neq N$
Thanks!
It's the definition of the product on the quotient group: by definition $$ (xN)(yN)=xyN $$ so there's really nothing to prove.
The neutral element in the quotient group is $1N=N$. So, by definition, $(gN)^0=N$, as $x^0$ is defined, in every group, to be the neutral element.
It is actually the case that if you consider that as the product set, under the convention that $X\cdot Y=\{xy:x\in X,y\in Y\}$, the equality $$ xN\cdot yN=(xy)N $$ holds. Indeed, since $N$ is normal, $Ny=yN$ and moreover $N\cdot N=N$ as $N$ is a subgroup. However, this is not really relevant for the definition of a group structure on $G/N$.