Given a group $G$, the automorphism group $Aut(G)$ of $G$ and the permutation group $\sum_G$ of $G$, I have to prove, that $Aut(G)$ is a subgroup of $\sum_G$.
To do that I must prove:
- $Aut(G)$ has exactly one neutral element.
- For each element $\phi$ in $Aut(G)$ there is an inverse element $\phi^{-1}$ in $Aut(G)$
- Associativity: $\phi\circ (\psi\circ\sigma )=(\phi\circ\psi )\circ\sigma$
- Closure under composition: If $\phi$ and $\psi$ are automorphisms in $Aut(G)$, then $\phi\circ\psi$ is also an automorphism in $Aut(G)$
Proof:
The identity map $Id: G\rightarrow G, g\mapsto g$ lies in $Aut(G)$ and is therefore the identity element.
Each automorphism $\phi$ in $Aut(G)$ has, by definition, an inverse map $\phi^{-1}$ which also lies in $Aut(G)$. Therefore each element has its inverse.
- The composition of maps is always associative. (Is this sufficient, or do I have to write more here?)
- Here, I will use the definition of a homomorphism: $\phi (x\circ y)=\phi (x)*\phi(y):$
$\forall g,h\in G,\forall\phi ,\psi\in Aut(G):(\phi\psi)(gh)=\phi(\psi(gh))=\phi(\psi(g)\psi(h))=\phi(\psi(g))\phi(\psi(h))=(\phi\psi)(g)(\phi\psi)(h)$
Is this proof correct? Also, it seems to me that it is a little bit too long. Are there any things which I can omit? I think 4. is the significant point. Would it suffice if I write just that and leave the other things as implied?