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I'm having difficulties understanding the definition of the right regular representation as it appears in Dummit & Foote's Abstract Algebra text. On page 132 it says

Let $\pi:G \to S_G$ be the left regular representation afforded by the action of $G$ on itself by left multiplication. For each $g \in G$ denote the permutation $\pi(g)$ by $\sigma_g$, so that $\sigma_g(x)=gx$ for all $x \in G$. Let $\lambda:G \to S_G$ be the permutation representation afforded by the corresponding right action of $G$ on itself, and for each $h \in G$ denote the permutation $\lambda(h)$ by $\tau_h$. Thus $\tau_h(x)=xh^{-1}$ for all $x \in G$ ($\lambda$ is called the right regular representation of $G$).

I can't make since of that definition. Earlier, on page 129 the authors explain exactly what is the right action corresponding to some given left action:

For arbitrary group actions it is an easy exercise to check that if we are given a left group action of $G$ on $A$ then the map $A \times G \to A$ defined by $a \cdot g=g^{-1} \cdot a$ is a right group action. Conversely, given a right group action of $G$ on $A$ we can form a left group action by $g \cdot a=a \cdot g^{-1}$. Call these pairs corresponding group actions.

If I try to find the right group action corresponding to $g \cdot a=ga$, I get $a \cdot g:=g^{-1} \cdot a=g^{-1}a$. Hence it seems to me that the definition should be $\tau_h(x)=h^{-1}x$ and not $xh^{-1}$.

Are there any flaws with my reasoning?

Thanks!

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  • $\begingroup$ $gG=G=Gg$, in first equality we see the permutation effect of $g$ as a left multiplication and in second equality we see the effect as an right multiplicatin. I think that is the main differences between right and left action. putting $g^{-1}$ etc has notational advantage, I guess. $\endgroup$ – mesel Jul 14 '14 at 13:16
  • $\begingroup$ @mesel while I make sense of your comment, I still can't see why the definition of the right regular representation involves multiplication on the right. If right actions are associated to left actions as the authors claim it doesn't seem right. $\endgroup$ – user1337 Jul 19 '14 at 14:17
  • $\begingroup$ I think the word "corresponding" in the first quoted paragraph must be removed. There is no general way in which $\pi$ and $\lambda$ can be obtained one from another without knowing that we are acting on $G$ itself. You might want to point this out to Richard Foote ( cems.uvm.edu/~rfoote ), who lists the errata for the book on his webpage. $\endgroup$ – darij grinberg Jul 30 '14 at 11:32
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In general, if $G$ acts on a set $S$ from the right via $(x,g) \mapsto x \cdot g$, then we obtain an action of $G$ on $S$ from the left via $g \cdot x := x \cdot g^{-1}$.

Let $*$ be the group multiplication in $G$. Then $G$ acts on $|G|$ (the underlying set of $G$, don't confuse it with $G$) from the right via $x \cdot g := x * g$. It follows that we obtain a left action from $G$ on $|G|$ via $g \cdot x := x \cdot g^{-1} = x * g^{-1}$. And this corresponds to a homomorphism of groups $G \to S(|G|)$ mapping $g$ to the permutation $(x \mapsto x * g^{-1}$).

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  • $\begingroup$ Thank you for your answer, but my question remains unclear. The authors refer to $g \cdot x=x*g^{-1}$ as a right group action. How can this be the action corresponding to the one given by left multiplication? $\endgroup$ – user1337 Jul 26 '14 at 17:45
  • $\begingroup$ @user1337 Wrong, the author (in the second blockquote of your question) quite plainly call $g\cdot x=x\cdot g^{-1}$ a left group action. Read it! $\endgroup$ – blue Jul 31 '14 at 9:40
  • $\begingroup$ @blue I was referring to the first blockquote: "...afforded by the corresponding right action...". $\endgroup$ – user1337 Jul 31 '14 at 9:52
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You seem to worry about these sentences:

"Let $\lambda : G \to S_G$ be the permutation representation afforded by the corresponding right action of $G$ on itself, and for each $h \in G$ denote the permutation $\lambda(h)$ by $\tau_h$. Thus $\tau_h(x)=xh^{−1}$ for all $x \in G$ ($\lambda$ is called the right regular representation of $G$)."

What you did at the end of your post is the following: You considered the left group action of $G$ on itself by left multiplication. Then you constructed the corresponding right group action and what you get is, correctly, $x \leftharpoonup h = h^{-1} x$.

The authors do something else in the above paragraph. They first consider the right action of $G$ on itself by right multiplication, i.e. $x \leftharpoonup g := xg$. But they want to obtain a "permutation representation", which they denote by $\lambda : G \to S_G$. Usually one wants this map to be a group homomorphism. In order to assure this, one needs to start with a left group action. What one can do is therefore the following: Consider the left group action of $G$ on itself corresponding to the above right group action of $G$ on itself by right multiplication. By your second quote this is given by $g \rightharpoonup x = x \leftharpoonup g^{-1} = x g^{-1}$. And this is exactly what the authors claim.

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  • $\begingroup$ I'm sorry for not getting this, but in the blockquote I can't find any hints for a left action. $\endgroup$ – user1337 Jul 31 '14 at 15:04
  • $\begingroup$ The point is that you want that the map $\lambda : G \to S_G$ to be a morphism of groups. Given an arbitrary left G-action on a set X, you get a morphism of groups $G \to S_X$ by sending $g$ to the map $X \to X, x \mapsto (g \rightharpoonup x)$. If however you have a right action of $G$ on $X$, then $G \to S_X, g \mapsto x \mapsto (x \leftharpoonup g)$ does not define a morphism of groups. $\endgroup$ – user44400 Jul 31 '14 at 16:31

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