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p. 6: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf enter image description here

Pretend the blue set was not given and I have to calculate it myself:
For all $x \in G, f(x) = xgH$ is faithful $\iff \color{blue}{\text{some set which I must find}} = \color{green}{\{id\}} $.

The PDF unfolds the answer: Because $ {g_2}^{-1}g_1 \in \color{blue}{\bigcap_{g \in G} gHg^{-1}} $, hence $ \color{blue}{\bigcap_{g \in G} gHg^{-1}} = \color{green}{\{id\}} $.

I tried: To determine the values of $ x $ for which $ f(x) = xgH$ is faithful, I solve $ g_1 * x = g_2 * x $. Here I let * represent the group binary operation.

$ g_1 * x = g_2 * x :\iff g_1(gH) = g_2(gH) \text{ for all } g \in G \iff g_2^{-1}g_1 \in \color{red}{gH}$.
The last $\iff$ is by dint of the result: ${ aH = bH \iff b^{-1}a \in H} $.

What did I bungle? I missed $\color{red}{\bigcap_{g \in G}}$ and $g^{-1}$?

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    Honestly I cannot follow your argument at all. What is $f$? Are $\{g_i\}$ some system of coset representatives? What does $g_1x=g_2x$ have to do with anything? Where is $x$ in the equation $f(g_i)=g_igH$? Note that $aH=bH\iff b^{-1}aH$ does **not** mean that $(g_1gH=g_2gH\forall g\in G)\implies g_2^{-1}g_1\in gH$ (this proposition doesn't even make sense, since $g$ is varying over $G$, and so the cosets are varying, whereas $g_2^{-1}g_1$ is fixed wrt $g$).2012-12-25

3 Answers 3

1

The action is faithful iff

$\left(\forall\,x\in G\,\,,\,g(xH)=xH\Longleftrightarrow g=1\right)\Longleftrightarrow \left(\forall\,x\in G\,\,,\,gxH=xH\Longleftrightarrow g=1\right)$

$\Longleftrightarrow \,\,\left(\forall\,x\in G\,\,,\,x^{-1}gx\in H\Longleftrightarrow g= 1\right)\,\Longleftrightarrow \,\,\forall\,x\in G\,\,,\,g\in xHx^{-1}=1$

5

The group action is faithful if the action of an element $x$ is trivial iff $x=1_G$. That is, if

$(xgH=gH\text{ for each }g\in G)\iff x=e $

Of course, the condition $xgH=gH$ can be rewritten as

$g^{-1}xgH=H\iff g^{-1}xg\in H\iff x\in gHg^{-1}.$

Is the path forward clear now?

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    @Frank: How can you have $\forall x$ on the left of the iff sign, and another $x$ on the right? *What does that even mean?* How can $b^{-1}a\in x_1H$ and $b^{-1}aH\in x_2H$ when (say) $x_1H$ and $x_2H$ happen to be disjoint cosets (answer: it's impossible). It further doesn't make sense to ask "for what values of $x$" makes a $\forall x$ statement true. Please reflect on what "for all $x$" means. In the statement $aH=bH\iff b^{-1}a\in H$, you cannot replace the subgroup $H$ with some coset $xH$ of $H$ - why are you convinced that works? Do you know *why* $b^{-1}aH=H$ implies $b^{-1}a\in H$?2012-12-26
4

It really doesn’t make sense to talk about the values of $x$ for which the action is faithful: faithfulness is a property of the action as a whole. I can’t make any sense of $\implies g_2^{-1}g_1\;:$ $g_2^{-1}g_1$ isn’t something that can be implies. It’s an element of the group $G$, not a statement.

Suppose that $\bigcap_{g\in G}gHg^{-1}=\{1_G\}$. In order to show that the action is faithful, you must show that if $g_1,g_2\in G$, and $g_1gH=g_2gH$ for all $g\in G$, then $g_1=g_2$. Now

$\begin{align*} g_1gH=g_2gH\quad&\text{ iff }\quad g^{-1}g_2^{-1}g_1g\in H\\ &\text{ iff }\quad g_2^{-1}g_1\in gHg^{-1}\;, \end{align*}\tag{1}$

so if $g_1gH=g_2gH$ for all $g\in G$, we must have

$g_2^{-1}g_1\in\bigcap_{g\in G}gHg^{-1}\;.$

If $\bigcap_{g\in G}gHg^{-1}=\{1_G\}$, this implies that $g_2^{-1}g_1=1_G$ and hence that $g_1=g_2$, as desired.

Conversely, suppose that $\bigcap_{g\in G}gHg^{-1}\ne\{1_G\}$, and fix $g_1\in\bigcap_{g\in G}gHg^{-1}$ with $g_1\ne 1_G$. Let $g_2=1_G$; you can reverse the calculations above to show that $g_1gH=g_2gH$ for all $g\in G$ and hence that the action is not faithful: $g_1$ and $1_G$ act identically on $G/H$.

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    @Frank: You’re very welcome. I’m glad that you got it sorted.2012-12-25