In proving the the set $Aut(G)$ is an automorphism, we set out to prove that the automorphism forms a group under functional composition.
Closure, associativity, existence of inverse function doesn't seem to hard.
Inverse function is what I find myself having issues with.
suppose $\alpha, \beta$ is an automorphism
$\alpha$, $\beta : G\rightarrow G$
Indeed, $\alpha$ is an automorphism implies it is bijective. The inverse of a bijective function is bijective.
To show that $\alpha$ is an automorphism, we seek to show that $\alpha$ is a an isomorphism $\alpha$:$ G\rightarrow G$
and if we can prove this, we are done.
Here is where I run into a brick wall. Referring to my notes,
it states that
$\forall x_{1},x_{2} \in G \exists g_{1},g_{2} \in G s.t g_{1}=\left ( x_{1} \right )\alpha^{-1}$ and $g_{2}=\left ( x_{2} \right )\alpha^{-1}$ If and only If $\left ( g_{1} \right )\alpha=x_{1}$ and$ \left ( g_{2} \right )\alpha=x_{2}$
Is the above necessart to prove homomorphism?
Best Answer
To prove that the inverse $\alpha^{-1}$ is an automorphism (apart from bijectivity) you need to establish $$\forall y_1,y_2 \in G \quad \alpha^{-1}(y_1y_2)=\alpha^{-1}(y_1) \, \alpha^{-1}(y_2).$$ Since $\alpha$ is onto, therefore there exists $x_i \in G$ such that $\alpha(x_i)=y_i$. Therefore, \begin{align*} \alpha^{-1}(y_1y_2) & = \alpha^{-1}(\alpha(x_1) \, \alpha(x_2))\\ & = \alpha^{-1}(\alpha(x_1x_2))\\ & = x_1x_2\\ &=\alpha^{-1}(y_1) \alpha^{-1}(y_2). \end{align*}