The following corollary is quite confusing to me.
Let $N$ be a normal subgroup of a group, $G$ and let $\bar{G}$ denote the set of cosets of $N$ in $G$. Let $\pi: G \rightarrow \bar{G}$ be the canonical homomorphism. Let $a_{1},...a_{k}$ be elements of $G$ such that the product $a_{1}...a_{k}$ is in $N$. Then $\bar{a_{1}}...\bar{a_{k}} = 1$.
The proof is that the product $p=a_{1}...a_{k}$ is in $N$, so $\pi(p) = \bar{p} = \bar{1}$. Since $\pi$ is a homomorphism, so $\bar{a_{1}}...\bar{a_{k}} = \bar{p}$.
I'm wondering why it equals 1.
NB: a bar over something indicates an element from the set of cosets... I think (ref. page 66, Algebra by Artin).
Thanks in advance for your help.