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let $X$ be a topological space on which a group $G$ acts.

1) is it true that this action always induces an homomorphism $G\rightarrow Aut(X)$?

My guess is no. because i think the induced homomorphism would be the map that associates to $g\in G$ the map $\phi_g:X\rightarrow X; \, x\mapsto gx$. Here $\phi_g$ is clearly injective but it need not be surjective unless the action is transitive. Hence $\phi_g\not \in Aut(X)$.

2)Is it true that if $G$ is finite then the action is always transitive?

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    1) $(\phi_{g})^{-1} = \phi_{g^{-1}}$ 2) No. Take a trivial action.2011-07-03
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    You misunderstand the definition of transitivity.2011-07-03
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    @Theo Buehler : thanks !! I see my error now in 1). in 2) i meant a non trivial action of course.2011-07-03
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    (1) Just to make Theo's first commet a bit more explicit, consider that $g$ sends $g^{-1}x$ to $x$. (2) Whether you actually get an homomorphism or an anti-homomorphism may depend on your conventions on left or right actions (if you find this second comment confusing, just ignore it)2011-07-03
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    Still, why should it be? Take a set $X$ of larger cardinality than $G$ then the action cannot possibly be transitive. E.g. take $G \cup \{pt\}$ with left multiplication action of $G$ on itself.2011-07-03
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    @Theo Buehler: the converse of 1) is obviously true i mean every such homomorphism defines an action of $G$ on $X$. how can we formulate this, can we say that the set $Hom(G,Aut(X))$ is in one to one correspondance with the set of actions $G\times X\rightarrow X$ ?2011-07-03
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    Yes, that's true. Given an action $\alpha: G \times X \to X$ the map $g \mapsto \phi_g = \alpha(g,\cdot)$ is a homomorphism $\phi: G \to \operatorname{Aut}{(X)}$ (by the axioms imposed on $\alpha$). Conversely, given a homomorphism $\phi: G \to \operatorname{Aut}{(X)}$ put $\alpha(g,x) = [\phi(g)](x)$. You should check that these two maps are inverses of each other.2011-07-03
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    @Theo Buehler: which maps are inverse to each other? do you mean $\alpha$ and $\phi$? the domain of one is different from the range of the other!! or may be you mean the map that sends $\alpha$ to $\phi$ is inverse to the map sending $\phi$ to $\alpha$2011-07-03
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    The latter... $\alpha \mapsto \phi$ and $\phi \mapsto \alpha$.2011-07-03
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    Someone should turn all these comments into an answer...2011-07-03

1 Answers 1

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Following Mariano's suggestion, I'm turning my comments into an answer.

Recall that a left action of $G$ on $X$ can be described by a map $\alpha: G \times X \to X$ satisfying

  1. $\alpha(e,x) = x$ for all $x \in X$ (the neutral element acts trivially)
  2. $\alpha(g,\alpha(h,x)) = \alpha(gh,x)$ for all $g,h \in G$ and all $x \in X$ "associativity".

Combining 1. and 2. gives in particular $\alpha(g^{-1},\alpha(g,x)) = \alpha(e,x) = x = \alpha(g^{-1},\alpha(g,x))$. Writing $\phi_{g} = \alpha(g,\cdot)$ we see that $\phi_{g^{-1}} \circ \phi_{g} = \operatorname{id}_{X} = \phi_{g} \circ \phi_{g^{-1}}$, so $\phi_{g}: X \to X$ is a bijection of $X$ with inverse $\phi_{g^{-1}}$, no matter if the action is transitive or not. Moreover, $\phi_{gh} = \phi_{g}\circ\phi_{h}$ follows immediately from 2. and means that $\phi: G \to \operatorname{Aut}{(X)}$ is a homomorphism.

On the other hand, given a homomorphism $\phi: G \to \operatorname{Aut}{(X)}$ we get an action $\alpha$ by setting $\alpha(g,x) = \phi_{g}(x)$. It is not difficult to check that the correspondence $\alpha \leftrightarrow \phi$ gives mutually inverse bijections between the set of actions $G \times X \to X$ and the set of homomorphism $G \to \operatorname{Aut}(X)$.

As for the second question, there is no reason for a finite group to act transitively on a set. It can't act transitively if $|G| \lt |X|$, so for instance letting $X = G \cup \{pt\}$ where $G$ acts on itself by left translations and fixes the point $pt$, we have an action that isn't transitive.