Orbit, stabilizer and fixed points of a group's action on left cosets by left multiplication

Question

For this example - with hardly any steps - I'm not getting its conclusions with my work. Can someone please see where I might have gone wrong? I added the colors. Thank you.

Given Example

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

1. There is one orbit because $\color{red}{g \cdot H = gH \in \mathrm{Orb}_{\{H\}}} $.

2. We can find a $ g \in G$ which will map $ x $ to any point in $ G$. Therefore the orbit of any left coset is the whole set $ G/H $.

3. $\mathrm{Stab}_{xH} = \{g : gxH = xH \} = \{g : x^{-1}gx \in H\} = \color{blue}{xHx^{-1}} $

4. There are no fixed points if $H \neq G$.

What I tried

For 1. and 2.,

$$\mathrm{orb}_{xH} := \{g \cdot xH : g\in G\} = \color{brown}{\{gxH : g \in G\}}.$$ $$\because g,x \in G \therefore gx \in G \text{ so } \color{brown}{\{gxH : g \in G\}} = \{(gx)H : g \in G\} = G/H .$$ I understand $ \color{red}{\mathrm{Orb}_{\{H\}} := \{g \cdot \{H\} : g \in G\}} $. But how is this the same as the 2 sentences above?

3.$\text{}$ How does $ \{g \in G : x^{-1}gx \in H\}= \color{blue}{xHx^{-1}} $? I understand $ \color{blue}{xHx^{-1} = \{h \in H : xhx^{-1}\}}$

4.$\text{}$ I know the definition for $x$ to be a fixed point ($\color{green}{g \cdot x = x\; \forall g \in G}$), but how do you determine the fixed points methodically? Do I solve for $x$ in $\color{green}{g \cdot x = x}$?

I also tried this: 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 $\color{green}{\text{definition of a fixed point}}$?

Answer 1

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.

Answer 2

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.