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I have been studying group actions for about a month, so I am at a basic stage and forgive me if this turns out to be a stupid question. It is about the two equivalent definitions of group action and permutation representation. In the 'action' definition of a left group action, we use a map $ \alpha: G \times X \rightarrow X $, which I understand can be curried to give a chain of two functions $ \phi : G \rightarrow (X \rightarrow X) $, which leads to the alternative definition of the group action as a permutation representation $ \phi : G \rightarrow Sym(X)$.

But how does this work for the right group action $ \alpha: X \times G \rightarrow X $, as following the same currying technique would give a chain of functions $ \phi: X \rightarrow (G \rightarrow X)$ and not a permutation representation?

In addition, what are the main reasons for wanting two equivalent definitions as an action and permutation representation? It seems to me that the definition as a group representation is quite natural, whereas the action definition is axiomatic and seems to come from nowhere. Many thanks.

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Right action is precisely left action of the opposite group. –  Alexei Averchenko Nov 22 '12 at 17:37
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2 Answers 2

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Your observation is correct, indeed left group actions are exactly those functions corresponding to group homomorphisms (seen as functions from $G$ to the endofunction monoid $\text{End}(X)=\mathbf{Set}(X,X)$) via the following adjunction. $$\mathbf{Set}(G \times X, X) \cong \mathbf{Set}(G,\text{End}(X))$$

This correspondence extend to right group actions, but the object which correspond to these actions are the antihomomorphisms of groups.

In general an antihomomorphism between the two groups $G$ and $H$ is just a function $f \colon G \to H$ such that for each pair $x,y \in G$ the equality $$f(xy)=f(y)f(x)$$ holds. Another way to state this is to see an antihomomorphism as a group homomorphism from the opposite group of the $G$, namely $G^\text{op}$ to $\text{Sym}(X)$. $G^\text{op}$ is the group having as underlining set $G$ (so $G^\text{op}=G$ as set) having as multiplication the operation $\cdot \circ \sigma \colon G^\text{op} \times G^\text{op} \to G^\text{op}$ (where $\cdot \colon G \times G \to G$ is the group multiplication of $G$ and $\sigma \colon G \times G \to G \times G$ is the bijection sending the pair $(x,y)$ in the pair $(y,x)$, it's easy to prove that this is a group operation).

For every $r \colon X \times G \to X$ right action of $G$ on $X$ we can get the map $r \circ \sigma \colon G^\text{op} \times X = G \times X \to X$ which is a left action of the group $G^\text{op}$ on the set $X$. So by what you told in your answer this action via currying correspond to homomorphism $G^\text{op} \to \text{Sym}(X)$ which are antihomorphisms from $G$ to $\text{Sym}(X)$.

Hope the answer was sufficiently clear and helpful.

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Thanks for your answer, I think I will have to study more algebra in order to fully understand it (since I have just started at graduate level), but I think this concept of 'antihomomorphism' is what I was after, being the object corresponding to right actions. With regard to the second part of the question, which of the two definitions seems most natural to you (as an action, or homomorphism/antihomomorphism), or do they have different advantages? Thanks –  user50229 Nov 22 '12 at 18:08
    
After studying more carefully I found that this was a very good answer to my main question that I was looking for, thanks. –  user50229 Nov 23 '12 at 0:58
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@SJGreen you're welcome. About your last comment on the comparison between homomorphisms (but I prefer talk about representations) and actions: as you have already said they both have different advantages. Representations are good objects for theoretical reasons: they allows to deal with actions in terms of the objects of group theory (i.e. groups and group homomorphisms). This is useful because it enables to use general results about group to prove facts about actions and the other end. –  Giorgio Mossa Nov 23 '12 at 11:50
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Actions are less group theoretic than homomorphism (at least for what's my taste), anyway in practice actions arise more naturally than representations, and they gives us a incredible beautiful and useful symbolism to deal with representations (at this point is clear that studying representations of a group is the same as studying its actions). –  Giorgio Mossa Nov 23 '12 at 11:56
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These are just some few reasons which explain why both the definitions are important, probably there are many more which I've forgotten, but I'm sure you're going to get a better understanding of these relations and their importance in particular when you're going to start to deal with ring/algebras representations and modules over said algebras, which are the ring version of group representations and actions. –  Giorgio Mossa Nov 23 '12 at 11:58
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Those are not precisle equivalent definitions: any group action on a set determines a homomorphism from the group to the group of symmetries (permutations) of the set, and the other way around is also true.

Thus, if $\,\phi: G\times X\to X\,$ is a group action on a set $\,X\,$ (and it really doesn't matter whether this is a right or left action), then we get a homomorphism

$$\Phi:G\to\operatorname{Sym}_X\,\,\,,\Phi(g)(x):=\phi(g,x)$$

and this means that for any $\,g\in G\,$, the permutation $\,\Phi(g)\,$ is such that it maps and element $\,x\in X\,$ to the element $\,\phi(g,x)\,$ in $\,X\,$.

It's easy to check the above is really a groups homomorphism, and also that we can do the same thing the other way around: begin with a homom. $\,G\to\operatorname{Sym}_X\,$ and get an action $\,G\times X\to X\,$ uniquely determined by the homomorphism.

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Thanks for your answer, I was wondering do you have any insight into why an action is defined the way it is - which way of thinking about actions seems more natural to you, or do both alternatives have their advantages? –  user50229 Nov 22 '12 at 18:12
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