View source for Alg7.12

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

<math>1\isin N\,</math> because <math>N \le G\,</math><br><br>

Therefore <math>g=g\cdot 1 \isin gN \,\,\forall g\isin G\,</math> so <math>G=\bigcup_{g\isin G}gN\,</math>.<br><br>

To show that distinct left cosets are disjoint, suppose <math>uN\cap vN\ne \empty\,</math> and show that <math>uN = vN\,</math>.<br><br>

Let <math>x\isin uN\cap vN\,</math> Then <math>x=un=vm\,</math> for some <math>n,m\isin N\,</math>.<br><br>

Multiply both sides of the second equation by <math>n^{-1}.\,</math><br><br>

<math>u=vmn^{-1}=vm_1\,</math> with <math>m_1 =mn^{-1} \isin N\,</math><br><br>

Now I can write <math>u\,</math> as <math>vM\,</math>, so for some <math>p\isin N\,</math>,<br><br>

<math>up = (vm_1)p = v(m_1p) \isin vN\,</math><br><br>

This means <math>uN \subseteq vN\,</math>.<br><br>

Similarly <math>vN \subseteq uN\,</math> and so <math>uN=vN\,</math>. This finishes the first part of the proof.<br><br>

To show  <math>\forall u,v,\isin G, u N = v N\,</math> if and only if <math>v^{-1}u\isin N\,</math>,<br><br>

<math>uN=vN\implies u\subseteq vN \,</math> which means <math>u = vn\,</math> for some <math>n\isin N\,</math>.<br><br>

<math>u=vn\implies v^{-1}u=n\implies v^{-1}u\isin N\,</math><br><br>

Since <math>u\subseteq vN\,</math>, <math>uN=vN\,</math> if and only if <math>u\,</math> and <math>v\,</math> are representatives of the same coset.<br><br>

[[Absract Algebra]]<br><br>

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