Exercise 1.6.20 (Automorphism group of $G$)

Proof. By definition $\mathrm{Aut}(G)\subset S_G$, where $S_G$ is the group of the permutations of $G$ under function composition, so its suffices to prove that $\mathrm{Aut}(G)$ is a subgroup of $(S(G),\circ )$.

• Consider ${\mathrm{Id}}_G:G\to G$ the map defined by ${\mathrm{Id}}_G(g)=g$ for all $g\in G$. Then ${\mathrm{Id}}_G\in S_G$, and for all $g,h\in G$, ${\mathrm{Id}}_G(\mathit{gh})=\mathit{gh}={\mathrm{Id}}_G(g){\mathrm{Id}}_G(h)$, thus ${\mathrm{Id}}_G\in \mathrm{Aut}(G)$ and $\mathrm{Aut}(G)\ne \varnothing$.

• If $\phi ,\psi$ are automorphisms of $G$, then $\phi \circ \psi :G\to G$ is bijective and for all $g,h\in G$,

  $$\begin{array}{llll}\hfill (\phi \circ \psi )(\mathit{gh})& =\phi (\psi (\mathit{gh}))& \hfill & \\ \hfill & =\phi (\psi (g)\psi (h))& \hfill & \\ \hfill & =\phi (\psi (g))\phi (\psi (h))& \hfill & \\ \hfill & =(\phi \circ \psi )(g)(\phi \circ \psi (h)),& \hfill & \end{array}$$

  thus $\phi \circ \psi \in \mathrm{Aut}(H)$.

• Moreover, ${\phi }^{-1}$ is bijective. Let $g,h$ be any elements in $G$. Put $a={\phi }^{-1}(g),b={\phi }^{-1}(h)$, so that $\phi (a)=g,\phi (b)=h$. Then

  $$\begin{array}{llll}\hfill \mathit{gh}& =\phi (a)\phi (b)& \hfill & \\ \hfill & =\phi (\mathit{ab}),& \hfill & \end{array}$$

  therefore

  $${\phi }^{-1}(\mathit{gh})=\mathit{ab}={\phi }^{-1}(g){\phi }^{-1}(h).$$

  This shows that ${\phi }^{-1}\in \mathrm{Aut}(G)$.

So $\mathrm{Aut}(G)$ is a subgroup of $(S(G),\circ )$, which means that $\mathrm{Aut}(G)$ is a group under function composition. □