I posted a question some time ago which was poorly received. Admittedly, my lecture notes were pretty sloppy too so that could have played a part in my inability to frame the question I wanted.
which obviously is sloppy...
Automorphism of a group is a group action
Here is the definition for group action:
Let G be a group, $\Omega$ be a finite set. A function $\mu: \Omega \times G\rightarrow \Omega$
is called an action of G on $\omega$ if two properties are satisfied:
1)$\mu \left ( \omega,e\ \right )=\omega$
2) $\mu\left ( \omega,gh \right )=\omega^{gh}=\mu\left ( \mu\left ( \omega,g \right ),h \right )$
Diving into the problem:
Given the definition for the action of a group G on a set, the fact that Aut(G) acts on $\Omega=G$ comes across as Aut(G) being the group action.
In the linked given, a poster has propositioned a map but going by the general definition of the map given by me above,
it would seem that the map is
$\mu:Aut\left ( G \right ) \times \Omega \rightarrow G$
$\left ( \phi,g \right ) \mapsto g $
so we ought to check for
$\mu\left ( \phi_{1}\phi_{2},g \right ) =g^{\phi_{1}\phi_{2}}$
and
$\mu\left ( \phi_{e},g \right )$ ,compatibility and identity, respectively.
Am I right?
Any help to lay my doubts to rest is greatly appreciated.
Define the mapping $f: (Aut(G), \circ) \times G \to G: (\varphi,g) \mapsto\varphi(g) = \varphi.g$
Clearly this mapping is well defined, since $Aut(G)$ is a function $G \to G$
If $f$ defines an action, then we must have that:
1) $\forall \varphi,\phi \in Aut(G), \forall g \in G: \varphi.(\phi.g) = (\varphi \circ \phi).g$
2) $\forall g \in G: 1_G.g = g$
So let's prove these two properties: Take $\varphi,\phi \in Aut(G)$, $g \in G$
1) $(\varphi \circ \phi).g = \varphi \circ \phi(g) = \varphi(\phi(g)) = \varphi.\phi(g) = \varphi.(\phi.g)$
2) $1_G.g = 1_G(g) = g$
Hence, $f$ defines a group action from $Aut(G)$ on $G$