The definition of the right regular representation 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!

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.
