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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.

from lecture notes 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.

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    $\begingroup$ Perhaps part of the confusion is that you've got a $G$ in your definition of a group action and a $G$ in your specific example, but they're playing different roles -- $Aut(G)$ is the '$G$' in the group action you're considering, while $G$ is the '$\Omega$'. Why not rewrite the definition of a group action using, say, $H$ and $\Omega$, and use that to investigate the $Aut(G)$, $G$ example? $\endgroup$ – aPaulT Jul 12 '17 at 10:10
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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 $G$

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  • $\begingroup$ Using $\phi$ and $\varphi$ to denote two different morphisms makes things really hard to read. $\endgroup$ – Nephry Jul 12 '17 at 13:48

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