Clarification on normal subgroups and quotient groups 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.
 A: Recall that the cosets of $N$ form a partition of $G$: for any $a,b\in G$, either $aN=bN$, or $aN\cap bN=\varnothing$.
Now let $p\in N$. $N$ is a subgroup of $G$, so $p^{-1}\in N$, and therefore $1=pp^{-1}\in pN\cap N$. Thus, $pN$ and $N=1N$ are not disjoint, so they must be equal: $pN=1N$, or, in the bar notation, $\bar p=\bar 1$.
A: (Presumably redundant response to questions in comments; Brian's answer already contains this)
$\bar a$ is by definition $aN$, the coset of $N$ containing $a$.
If $\bar a = \bar b$, then by definition $aN = bN$.
Guido's lemma is that $aN = bN$ if and only if $b^{-1}aN = N$ if and only if $b^{-1}a \in N$.
While simple, this is extremely useful.
Proof: If $aN = bN$ then every element of $aN$ is contained in $bN$, so in particular, $a\cdot 1 \in aN$ is also in $bN$, so there is some $n \in N$ such that $a \cdot 1 = b \cdot n$. Dividing on the left by $b$ gives $b^{-1} a = n \in N$ and $b^{-1} a N = n N = N$.  Conversely if $b^{-1}a \in N$, then there is some $n \in N$ such that $b^{-1} \cdot a = n$, and then $a \cdot 1 = b \cdot n$ so that $b$ is in both $aN$ and $bN$. Since cosets are either equal or disjoint, $aN = bN$. $\square$
