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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 $g$). –  anon Dec 25 '12 at 19:26
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up vote 4 down vote accepted

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: I started from it, but as I said at the end, the argument can be reversed to find it. To see how your calculation should have gone, look at $(1)$ above; I’m not sure what you were thinking when you concluded that $g_2^{-1}g_1\in gH$. –  Brian M. Scott Dec 25 '12 at 20:33
    
Thank you. Sorry for the confusion, but the question actually asks us to find $ \color{blue}{\bigcap_{g \in G} gHg^{-1}} $ on our own. Although it was quick to prove $ g^{-1}ug \in H \iff u \in gHg^{-1} \forall u \in G$, I don't see how you'd realize that you'd need this property? I understand (1) after looking at the answer and your work, but if I didn't know what the answer was, I wouldn't know to do $$ g^{-1}{g_2}^{-1}g_1g \in H \iff g_2^{-1}g_1 \in gHg^{-1}.$$ How would you know to do this in the forward step? –  Frank Muer Dec 25 '12 at 20:36
    
@Frank: I just asked myself what it meant to have $g_1gH=g_2gH$ for all $g\in G$ and discovered immediately that this was equivalent to $g^{-1}g_2^{-1}g_1g\in H$ for all $g\in G$. The $g^{-1}g_2^{-1}g_1g$ is a conjugate of $g_2^{-1}g_1$, and $H$ is normal, so I immediately think of moving the $g$ factors to the other side: $g_2^{-1}g_1\in gHg^{-1}$ for all $g\in G$. And that tells me that $g_1$ and $g_2$ act identically on $G/H$ iff $g_2^{-1}g_1$ is in the intersection of all the conjugates of $H$, which is a fairly natural object. –  Brian M. Scott Dec 25 '12 at 20:41
    
Thank you so much again. Two last inquiries: (3) What didn't make sense about $ \implies {g_2}^{-1}g_1 \in gH $? I'm just using $ aH = bH \implies b^{-1}a \in H $? (4) I still don't see how to realize that I need the intersection. I know the meaning of it --- we want to emphasize that $ {g_2}^{-1}g_1 \in \text{ all of the } gHg^{-1} $. But at the start, we already said $ \forall g \in G $. Isn't this sufficient? Why add on the intersection sign at the end? –  Frank Muer Dec 25 '12 at 21:04
    
@Frank: (3) I didn’t see the $\in gH$: in the original post the $g_2^{-1}g_1$ was at the end of short line and was followed by a full stop, so I took it to be the end of the sentence. (4) It’s not a matter of emphasizing anything: we’ve discovered that $g_1$ and $g_2$ act identically on $G/H$ iff $g_2^{-1}g_1\in gHg^{-1}$ for every $g\in G$, and $g_2^{-1}g_1\in\bigcap_{g\in G}gHg^{-1}$ is simply a shorter way of saying that $g_2^{-1}g_1\in gHg^{-1}$ for every $g\in G$. I’m really not sure what your confusion here is. –  Brian M. Scott Dec 25 '12 at 21:14
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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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Thanks anon. I fixed some notation. but for what values of $ x $ is it true that $axH = bxH \text{ for all } x \in G \iff b^{-1}a \in xH $? I thought $ aH = bH \iff b^{-1}a \in H $ implied this. –  Frank Muer Dec 25 '12 at 20:26
    
@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 $b^{-1}a\in H$? –  anon Dec 26 '12 at 1:14
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The action is faithful iff

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

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

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