Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 27: Proof of the Orbit-Stabilizer theorem

Let $G$ be a group and let $S$ be a $G\text{-set.}$

(a)	The orbits partition $S\text{.}$
(b)	If $s\in S$ and $H=\text{Stab}\left(s\right)$ then $$\begin{array}{cccc}\phi :& \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.& ⟶& Gs\\ & gH& ⟼& gs\end{array}$$ is a function and $\phi$ is a bijection.

Let $G$ be a group and let $N$ be a subgroup of $G\text{.}$

(a)	The cosets in $\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$N$}\right.$ partition $G\text{.}$
(b)	All cosets have the same size.

		Idea of proof.
	Let $g\in G\text{.}$ To show: $$\begin{array}{cccc}\phi :& gN& ⟶& N\\ & gn& ⟼& n\end{array}$$ is a function and $\phi$ is a bijection. $\square$

Let $\sim$ be an equivalence relation on a set $S\text{.}$

The equivalence classes partition $S\text{.}$

		Proof of the first Proposition.
	(a) To show: The orbits partition $S\text{.}$ To show: (aa) $\bigcup _{s\in S}Gs=S\text{.}$ (bb) If $s_1,s_2\in S$ and $G{s}_1\cap G{s}_2\ne \varnothing$ then $G{s}_1=G{s}_2\text{.}$ (aa) To show: (aaa) $\bigcup _{s\in S}Gs\subseteq S\text{.}$ (aab) $S\subseteq \bigcup _{s\in S}Gs\text{.}$ (aaa) Since $Gs\subseteq S$ then $\bigcup _{s\in S}Gs\subseteq S\text{.}$ (aab) To show: If $a\in S$ then $a\in \bigcup _{s\in S}Gs\text{.}$ Since $a\in S,$ and $a\in Ga,$ then $a\in \bigcup _{s\in S}Gs\text{.}$ So $S\subseteq \bigcup _{s\in S}Gs\text{.}$ (ab) Assume $s_1,s_2\in S$ and $G{s}_1\cap G{s}_2\ne \varnothing \text{.}$ To show: $G{s}_1=G{s}_2\text{.}$ Since $G{s}_1\cap G{s}_2\ne \varnothing$ there exists $t\in G{s}_1\cap G{s}_2\text{.}$ So there exist $g_1,g_2\in G$ such that $$g_1{s}_1=t=g_2{s}_2\text{.}$$ So $s_1=g_1^{-1}{g}_2{s}_2$ and $s_2=g_2^{-1}{g}_1{s}_1\text{.}$ To show: (aba) $G{s}_1\subseteq G{s}_2\text{.}$ (abb) $G{s}_2\subseteq G{s}_1\text{.}$ (aba) To show: If $\ell \in G{s}_1$ then $\ell \in G{s}_2\text{.}$ Assume $\ell \in G{s}_1\text{.}$ Then there exists $h\in G$ such that $\ell =h{s}_1\text{.}$ So $\ell =h{s}_1=h{g}_1^{-1}{g}_2{s}_2\in G{s}_2,$ since $h{g}_1^{-1}{g}_2\in G\text{.}$ So $G{s}_1\subseteq G{s}_2\text{.}$ (abb) To show: If $m\in G{s}_2$ then $m\in G{s}_1\text{.}$ Assume $m\in G{s}_2\text{.}$ Then there exists $k\in G$ such that $m=k{s}_2\text{.}$ So $m=k{s}_2=k{g}_2^{-1}{g}_1{s}_1\in G{s}_1,$ since $k{g}_2^{-1}{g}_1\in G\text{.}$ So $G{s}_2\subseteq G{s}_1\text{.}$ So $G{s}_1=G{s}_2\text{.}$ So the orbits partition $G\text{.}$ (b) To show: (ba) $\begin{array}{cccc}\phi :& \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.& ⟶& Gs\\ & gH& ⟼& gs\end{array}$ is a function. (bb) $\phi$ is a bijection. (ba) To show: If $g_{1}H,g_{2}H\in \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.$ and $g_{1}H=g_{2}H$ then $\phi \left(g_{1}H\right)=\phi \left(g_{2}H\right)\text{.}$ Assume $g_1,g_2\in G$ and $g_{1}H=g_{2}H\text{.}$ Then $g_1\in g_{2}H\text{.}$ So there exists $h\in H$ with $g_1=g_{2}h\text{.}$ To show: $\phi \left(g_{1}H\right)=\phi \left(g_{2}H\right)\text{.}$ To show: $g_{1}s=g_{2}s\text{.}$ $$g_{1}s=g_{2}hs=g_{2}s,$$ since $h\in \text{Stab}\left(s\right)\text{.}$ (bb) To show: $\phi$ is a bijection. To show: $\begin{array}{cccc}\psi :& Gs& ⟶& \raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.\\ & t& ⟼& gH\end{array}$ where $g\in G$ such that $t=gs$ is an inverse function to $\phi \text{.}$ To show: (bba) If $g_1,g_2\in G$ and $g_{1}s=g_{2}s$ then $\psi \left(g_{1}s\right)=\psi \left(g_{2}s\right)\text{.}$ (bbb) $\phi \circ \psi ={\text{id}}_{\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}$ and $\psi \circ \phi ={\text{id}}_{Gs}\text{.}$ (bba) Assume $g_1,g_2\in G$ and $g_{1}s=g_{2}s\text{.}$ Then $g_1^{-1}{g}_{2}s=s,$ so that $g_1^{-1}{g}_2\in \text{Stab}\left(s\right)\text{.}$ To show: $\psi \left(g_{1}s\right)=\psi \left(g_{2}s\right)\text{.}$ To show: $g_{1}H=g_{2}H\text{.}$ To show: (bbaa) $g_{1}H\subseteq g_{2}H\text{.}$ (bbab) $g_{2}H\subseteq g_{1}H\text{.}$ (bbaa) To show: If $x\in g_{1}H$ then $x\in g_{2}H\text{.}$ Assume $x\in g_{1}H\text{.}$ Then there exists $h\in H$ such that $x=g_{1}h\text{.}$ To show: $x\in g_{2}h\text{.}$ $$x=g_{1}h=g_2{g}_2^{-1}{g}_{1}h\in g_{2}H,$$ since $g_2^{-1}{g}_1\in \text{Stab}\left(s\right)=H\text{.}$ So $g_{1}H\subseteq g_{2}H\text{.}$ (bbab) To show: If $y\in g_{2}H$ then $y\in g_{1}H\text{.}$ Assume $y\in g_{2}H\text{.}$ Then there exists $k\in H$ such that $y=g_{2}k\text{.}$ So $$y=g_{2}k=g_1{g}_1^{-1}{g}_{2}k\in g_{1}H,$$ since $g_1^{-1}{g}_2\in \text{Stab}\left(s\right)=H\text{.}$ So $g_{2}H\subseteq g_{1}H\text{.}$ So $g_{2}H=g_{1}H\text{.}$ (bbb) To show: $\phi \circ \psi ={\text{id}}_{\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}$ and $\psi \circ \phi ={\text{id}}_{Gs}\text{.}$ If $g\in G$ then $$(\phi \circ \psi )\left(gH\right)=\phi \left(\psi \left(gH\right)\right)=\phi \left(gs\right)=gH$$ and $$(\psi \circ \phi )\left(gs\right)=\psi \left(\phi \left(gs\right)\right)=\psi \left(gH\right)=gs\text{.}$$ So $\phi \circ \psi ={\text{id}}_{\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}$ and $\psi \circ \phi ={\text{id}}_{Gs}\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 27 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 7, 2011.

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