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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!

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  • $\begingroup$ I can understand $G$ and $H$, but what is $N$? $\endgroup$ – Bernard Mar 1 '15 at 14:54
  • $\begingroup$ Sorry I confused notation $H = N$, there is no $H$ really $\endgroup$ – elfeck Mar 1 '15 at 14:56
  • $\begingroup$ In every group, we define $a^0 = e$, right? That is, an element to the $0$ power must be the identity element. Considering $gN$ as an element of the quotient group $G/N$, then $(gN)^0$ must be the identity element of $G/N$, which is ... ? $\endgroup$ – pjs36 Mar 1 '15 at 15:05
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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 $G/N$.

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  • $\begingroup$ Thank you with your edit that pretty much answers it $\endgroup$ – elfeck Mar 1 '15 at 15:04

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