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I have the following problem:

We fix a group G and a set X. I have just shown the following two statements.

  1. Let $a:G\times X\rightarrow X$ be a left action of G on X. Show that the map $b:X\times G\rightarrow X, b(x,g)=a(g^{-1},x)=g^{-1}x$ defines a right action of G on X

  2. Let $b:X\times G\rightarrow X$ be a right action of G on X. Show that the map $a:G\times X\rightarrow X, a(g,x)=b(x,g^{-1})=xg^{-1}$ defines a left action of G on X

And now I need to show the following:

Show that the constructions in 1.,2. are inverses of each other and define bijections between the sets of right and left actions of G on X.

I really don't understand what they want, so do i first need to show that $a(b(x,g))=b(x,g)$ and $(b(a(g,x))=a(g,x)$?

Furthermore I don't see why I need to define multiple bijections, not only one?

Can please someone give me a hint for both question and maybe explain better what I need to do?

Thank you!

user123234
  • 2,885

1 Answers1

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Given a group $G$ and a set $X$, you can consider the set $L$ of all left actions of $G$ on $X$, and similarly, you can consider the set $R$ for right actions. You have constructed maps $f: L \to R$ and $g: R\to L$, defined by $f(a(g,x)) = b(x, g^{-1})$ and similarly for $g$. Now it remains to show that $g \circ f$ is the identity map on $L$, i.e. it sends a left action to itself, and similarly, to show that $f\circ g$ is the identity map on $R$. Note that you are only constructing a bijection of sets, rather than an isomorphism of groups or an isomorphism of $G$-sets.

One thing that may be getting in the way of this is the terminology that is being used for this problem - you may be confusing the maps $f: L \to R$ and the group action $a: G \times X \to X$. To show, for example, that $g\circ f$ is the identity map on $L$, take any group action $a: G\times X \to X$, and show that for all $x \in X, g\in G$, that $g\circ f(a(g,x)) = a(g,x)$.

Sam K
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