# Alg7.12

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

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

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

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

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

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

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

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

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

This means $\displaystyle uN \subseteq vN\,$ .

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

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

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

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

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