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1) Prove that each left $G$-set $X$ can be turned into a right $G$-set by defining $x\sigma = \sigma^{-1}x$, and that every right $G$-set arises in this way.

I have the left action $X \times G \to X$ and $x\sigma = \sigma^{-1}x$. To show 1) I have to show associativity and the identity:

$x(\sigma\mu)=(\sigma\mu)^{-1}x=(\mu^{-1}\sigma^{-1})x = \mu^{-1}(\sigma^{-1}x)$ (as we have given a left-G-set) $= \mu^{-1}(x\sigma) = (x\sigma)\mu$.

What I don't understand here is, how do I show "that every right $G$-set arises in this way"?

2) Prove that every $G\times H$-set $X$ can be turned into a $G-H$-biset by putting $\sigma x=(\sigma,1)h$ and $x\rho = (1,\rho)=(1,\rho^{-1})x$ with $\sigma\in G, x \in X, \rho \in H$, and that every $G-H$-biset arises in this way.

A $G-H$-biset we defined: $X$ is a left-$G$-set and right-$H$-set with $\forall \sigma\in G, x\in X$ and $\rho\in H$: $(\sigma x)\rho = \sigma(x\rho)$.

Here I don't know how to do it. If I try to show the associativity, I simply don't know how to start...

I'll be happy about any help :) Best Sara

share|improve this question
    
the notions are dual and a group $G$ contains elements and their inverses so every right action is dual to a left action. –  palio Oct 14 '12 at 22:12
    
Regarding $\#1$, I think the question is asking you to prove the converse... given a right $G$-set, show us which left $G$-set it can be obtained from. –  Alexander Gruber Oct 14 '12 at 22:13
    
Thanks so far!! :) But what means being dual?.... Best, Sara –  Martha Oct 14 '12 at 22:59

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