Alg7.12

Let $N\,$ be any subgroup of the group $G\,$. The set of left cosets of $N\,$ in $G\,$ form a partition of $G\,$. Furthermore, $\forall u,v,\in G,uN=vN\,$ if and only if $v^{{-1}}u\in N\,$ and in particular, $uN=vN\,$ if and only if $u\,$ and $v\,$ are representatives of the same coset.

$1\in N\,$ because $N\leq G\,$

Therefore $g=g\cdot 1\in gN\,\,\forall g\in G\,$ so $G=\bigcup _{{g\in G}}gN\,$.

To show that distinct left cosets are disjoint, suppose $uN\cap vN\neq \emptyset \,$ and show that $uN=vN\,$.

Let $x\in uN\cap vN\,$ Then $x=un=vm\,$ for some $n,m\in N\,$.

Multiply both sides of the second equation by $n^{{-1}}.\,$

$u=vmn^{{-1}}=vm_{1}\,$ with $m_{1}=mn^{{-1}}\in N\,$

Now I can write $u\,$ as $vM\,$, so for some $p\in N\,$,

$up=(vm_{1})p=v(m_{1}p)\in vN\,$

This means $uN\subseteq vN\,$.

Similarly $vN\subseteq uN\,$ and so $uN=vN\,$. This finishes the first part of the proof.

To show $\forall u,v,\in G,uN=vN\,$ if and only if $v^{{-1}}u\in N\,$,

$uN=vN\implies u\subseteq vN\,$ which means $u=vn\,$ for some $n\in N\,$.

$u=vn\implies v^{{-1}}u=n\implies v^{{-1}}u\in N\,$

Since $u\subseteq vN\,$, $uN=vN\,$ if and only if $u\,$ and $v\,$ are representatives of the same coset.