Proving a set of automorphisms of a group is a group under composition

Let $G$ be a group and let $\mathrm{Aut}(G)$ be the set of all automorphisms of $G$. Prove that $\mathrm{Aut}(G)$ is a group under the operation of composition for functions.

I am not too sure if I am proving $\mathrm{Aut}(G)$ itself is a group. Would this mean then that I would have to make sure it fulfills all the group requirements? If so I am not too sure how to go about it. And to prove closure, I would think I need to pick say $f,g \in \mathrm{Aut}(G)$ then use the fact that these are automorphisms. From here, I am not too sure how to proceed.

Solution 1:

Let $G$ be a group. By definition $$\text{Aut}(G)=\left\{f:G\rightarrow G\mid f \mbox{ is an isomorphism of }G\right\}.$$ Given two automorphisms $f,g\in \text{Aut}(G)$, we can consider the composition $g\circ f$. We claim that $\text{Aut}(G),\circ$ is a group. We have to check all axioms.

First of all we need to show that $g\circ f$ is again an automorphism, i.e. a homomorphism that is bijective. Now since $g$ and $f$ are bijective, $g\circ f$ is bijective. Moreover,

$$\begin{align} (g\circ f)(ab)&=g(f(ab))\\ &=g(f(a)f(b))\\ &=g(f(a))g(f(b))\\ &=(g\circ f)(a)(g\circ f)(b), \end{align}$$

for all $a,b\in G$. Hence $g\circ f$ is a group homomorphism.

Secondly we need to show that $\circ$ is associative, i.e. $(h\circ g)\circ f=h\circ (g\circ f)$. Just evaluate both morphisms at $a\in G$ and see that both expressions coincide due to the associativity of $G$.

Thirdly we need to check that there is a neutral element for $\circ$. Clearly $Id_G:G\rightarrow G:a\mapsto a$ is an automorphism. Since $f\circ Id_G=Id_G\circ f$ for all $f\in \text{Aut}(G)$, $Id_G$ is the neutral element.

Last but not least, we have to check that each $f\in \text{Aut}(G)$ has an inverse for $\circ$. Consider the inverse function $f^{-1}$. Clearly $f^{-1}\circ f=Id_G=f\circ f^{-1}$. So it remains to show that $f^{-1}$ is a group morphism. Now it's a very good exercise to prove this last statement.

EDIT: Let's prove the last statement. Suppose that $f:G\rightarrow G$ is a group isomorphism. We need to show that $f^{-1}$ is a group morphism. Let $a,b\in G$. By definition there exist a unique $x,y\in G$ such that $f(x)=a$ and $f(y)=b$. Hence

$$\begin{align} f^{-1}(ab)&=f^{-1}(f(x)f(y))\\ &=f^{-1}(f(xy))\\ &=xy. \end{align}$$ Similarly

$$\begin{align} f^{-1}(a)f^{-1}(b)&=f^{-1}(f(x))f^{-1}(f(y))\\ &=xy. \end{align}$$

Hence $f^{-1}(ab)=f^{-1}(a)f^{-1}(b).$

Solution 2:

If we recognize that $\emptyset\ne\operatorname{Aut}(G)\subseteq\operatorname{Sym}(G)$, we're almost done. In fact, $\operatorname{Sym}(G)$ is a group$^{(1)}$, so we're just left to prove (by the subgroup lemma):

1. $\operatorname{Aut}(G)\ne\emptyset$;
2. $\varphi,\psi\in \operatorname{Aut}(G) \Rightarrow \varphi\psi\in \operatorname{Aut}(G)$;
3. $\varphi\in \operatorname{Aut}(G) \Rightarrow \varphi^{-1}\in \operatorname{Aut}(G)$.

Proof.

1. $\forall g,h \in G$: $$\iota_G(g)\iota_G(h)=gh=\iota_G(gh) \Rightarrow \iota_G \in \operatorname{Aut}(G) \Rightarrow \operatorname{Aut}(G)\ne\emptyset$$
2. $\forall g,h \in G$, $\forall \varphi,\psi \in \operatorname{Aut}(G)$: $$(\varphi\psi)(gh)=\varphi(\psi(gh))=\varphi(\psi(g)\psi(h))=\varphi(\psi(g))\varphi(\psi(h))=(\varphi\psi)(g)(\varphi\psi)(h) \Rightarrow \varphi\psi\in\operatorname{Aut}(G)$$
3. $\forall g,h \in G$, $\forall \varphi \in \operatorname{Aut}(G)$: $$gh=\iota_G(gh)=(\varphi\varphi^{-1})(gh)=\varphi(\varphi^{-1}(gh))$$ but also: $$gh=\iota_G(g)\iota_G(h)=(\varphi\varphi^{-1})(g)(\varphi\varphi^{-1})(h)=\varphi(\varphi^{-1}(g))\varphi(\varphi^{-1}(h))=\varphi(\varphi^{-1}(g)\varphi^{-1}(h))$$ whence ($\varphi$ is, in particular, injective): $$\varphi^{-1}(gh)=\varphi^{-1}(g)\varphi^{-1}(h), \space \forall g,h \in G \Rightarrow \varphi^{-1} \in \operatorname{Aut}(G)$$

$\Box$

Therefore, $\operatorname{Aut}(G)\le \operatorname{Sym}(G)$ and, in particular, $\operatorname{Aut}(G)$ is a group.

$^{(1)}$Here there's nothing to prove: the definition of the algebraic structure named "(abstract) group" is precisely patterned upon the properties of the set of the bijections on a given set, endowed with the composition of maps as internal operation (associativity, unit, inverses). So, in a sense, $\operatorname{Sym}(G)$ is the prototypical group.