Alg7.12

From Exampleproblems

Jump to: navigation, search

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

$1\isin N\,$ because $N \le G\,$

Therefore $g=g\cdot 1 \isin gN \,\,\forall g\isin G\,$ so $G=\bigcup_{g\isin G}gN\,$ .

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

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

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

$u=vmn^{-1}=vm_1\,$ with $m_1 =mn^{-1} \isin N\,$

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

$up = (vm_1)p = v(m_1p) \isin vN\,$

This means $uN \subseteq vN\,$ .

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

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

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

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

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

Absract Algebra

Retrieved from "http://www.exampleproblems.com/wiki/index.php/Alg7.12"

Views

Personal tools

Create an account or log in

Search

Toolbox

Argan Oil
Natural Skin Care
Organic Skin Care