# induced homomorphism from a group action

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?

• 1) $(\phi_{g})^{-1} = \phi_{g^{-1}}$ 2) No. Take a trivial action. – t.b. Jul 3 '11 at 16:49
• (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) – Andrea Mori Jul 3 '11 at 16:58
• 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. – t.b. Jul 3 '11 at 16:58
• 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. – t.b. Jul 3 '11 at 17:11
• The latter... $\alpha \mapsto \phi$ and $\phi \mapsto \alpha$. – t.b. Jul 3 '11 at 17:28

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.