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Reference: p. 8

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This PDF doesn't unfold all the steps hence can someone please notify me of bungles? Thank you.

I tried: $G$ acts on the left cosets of $H$ by left multiplication, so $ g \cdot xH = gxH $

(1.) $\mathrm{orb}_{xH} := \{g \cdot xH : g\in G\} = {\{gxH : g \in G\}}.$ $\because g,x \in G \therefore gx \in G \text{ hence } = \{(gx)H : g \in G\} = G/H .$

I understand $ \color{red}{\mathrm{Orb}_{\{H\}} := \{g \cdot \{H\} : g \in G\}} $. But how is my work overhead?

3.$\text{}$ How does $\mathrm{Stab}_{aH} = \{g : gaH = aH \} = \{g : a^{-1}ga \in H\} = \color{blue}{aHa^{-1}} $

I understand: $ \color{blue}{aHa^{-1} = \{h \in H : aha^{-1}\}}$

4.$\text{}$ How do you determine the fixed points craftily? I know the definition for $x$ to be a fixed point: $g \cdot x = x \; \forall g \in G$. Do I solve for $x$?

To boot, I tried from (3.) $g \cdot xH = xH \iff x^{-1}gx \in H \iff x \in H \text{ and } g \in H \iff G = H. $
But $g \cdot xH = xH $ isn't the definition of a fixed point?

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I'd recommend using markdown formatting for anything that isn't math. Also, I don't think the color formatting is necessary (and I find it more difficult to read when it is used so liberally). – Zev Chonoles Dec 26 '12 at 21:18

For $1$ and $2$ take $g=ax^{-1}$ for whichever $a\in G$ you want.

For $3$, $\{g : x^{-1}gx\in H \} = \{g : gx\in xH \} =\{g : g\in xHx^{-1} \}=xHx^{-1}$.

For $4$, if $H\not= G$, we can find an $a\notin H$ and look at the coset $aH$. Then $a^{-1}$ sends $aH$ to $H$, and similarly $a$ sends $H$ to $aH$, so no coset can be fixed under the action of $G$.

Another alternative way to see $4$ would be to use $3$. Take any coset $xH$, and we have that $\text{Stab}_G(xH)=xHx^{-1}$. Since $xHx^{-1}$ has order $|H|$, if $|H|<|G|$, then $$|\mathcal{O}_{xH}|=[G:\text{Stab}_G(xH)]=[G:H]>1$$ so $xH$ is not a fixed point.

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  1. My knowledge of group action is a bit rusty. My best hint for now is to look at what you want to prove: $gH \in \mathrm{Orb}_{\{H\}}$. You state that ${\mathrm{Orb}_{\{H\}} := \{g \cdot \{H\} : g \in G\}}$. What does this mean? More precisely, how can you use this definition to show that $gH \in \mathrm{Orb}_{\{H\}}$?

  2. Given an $x$, can you find the $g \in G$ which will map $x$ to a point in $G$? (Sorry that I'm not much help here. I know I'm just restating the question. This is exactly how my brain works trying to find a proof.)

  3. I'd step away from manipulating the sets themselves and chose an arbitrary element from $K = \mathrm{Stab}_{xH}$ and then show that it must be in $xHx^{-1}$ (and vice versa). So if $g \in K$ then there is some $x \in G$ such that $gxH = xH$. So there are $h, h' \in H$ such that $gxh = xh'$. So $g = xh'h^{-1}x^{-1} \in xHx^{-1}$. Can you do the converse from there?

  4. I'd start by assuming that there is a fixed point in $G$ if $H \not= G$. Try to find a contradiciton.

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I think you should try all this in a specific example, for example the symmetric group $S_3$ with presentation $ (x,y|x^2=y^3=xyxy=1 ) $ and the subgroup $H$ first the group generated by $x$ and then the group generated by $y$. – Ronnie Brown Dec 26 '12 at 22:17

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