The Group of Automorphisms of a Group, Aut(G)

# The Group of Automorphisms of a Group, Aut(G)

Recall from the Group Automorphisms page that if $(G, *)$ is a group then an automorphism on the group $(G, *)$ is a bijection $f : G \to G$ such that for all $x, y \in G$ we have that $f(x * y) = f(x) * f(y)$, i.e., an automorphism on $(G, *)$ is an isomorphism from the group $(G, *)$ to itself.

Now consider the set of automorphisms of the group $(G, *)$ which we denote by

(1)\begin{align} \quad \mathrm{Aut} (G) = \{ f : f : G \to G \: \mathrm{is \: bijective \: and \: for \: all} \: x, y \in G \: , f(x * y) = f(x) * f(y) \} \end{align}

In the following theorem we will prove that $(\mathrm{Aut}(G), \circ)$ forms a group.

Theorem 1: If $(G, *)$ is a group then $(\mathrm{Aut}(G), \circ)$ is also a group. |

**Proof:**Let $(G, *)$ be a group. Note that every element $f \in \mathrm{Aut}(G)$ is a bijection and the set of all bijections (permutations) is $S_G$, so clearly $\mathrm{Aut}(G) \subseteq S_G$. We have already proven that $(S_G, \circ)$ is a group, and so we only need to show that $\mathrm{Aut}(G)$ is closed under $\circ$ and that for all $f \in S_G$ there exists an $f^{-1} \in S_G$ such that $f \circ f^{-1} = \epsilon$ and $f^{-1} \circ f = \epsilon$ where $\epsilon$ is the identity permutation/bijection/automorphism.

- Let $f, g \in \mathrm{Aut} (G)$. Then $f, g : G \to G$ are bijections such that for all $x, y \in G$ we have that $f(x * y) = f(x) * f(y)$ and $g(x * y) = g(x) * g(y)$. Consider the composition $f \circ G$. We have:

\begin{align} \quad (f \circ g)(x * y) = f(g(x * y)) = f(g(x) * g(y)) = f(g(x)) * f(g(y)) = (f \circ g)(x) * (f \circ g)(y) \end{align}

- Therefore $(f \circ g) \in \mathrm{Aut} (G)$ so $\mathrm{Aut} (G)$ is closed under $\circ$.

- Now let $f \in \mathrm{Aut} (G)$. We know that since $\mathrm{Aut} (G) \subseteq S_G$ that $f^{-1}$ exists and $f^{-1} \in S_G$, so we only need to show that $f^{-1}$ is an automorphism to guarantee that $f^{-1} \in \mathrm{Aut} (G)$. Since $f : G \to G$ is a bijection we clearly have that $f^{-1} : G \to G$ is a bijection. Let $c = f(x), d = f(y) \in G$ so that $x = f^{-1} (c)$ and $y = f^{-1} (d)$. Then:

\begin{align} \quad f^{-1} (c * d) = f^{-1}(f(x) * f(y)) = f^{-1}(f(x * y)) = x * y = f^{-1}(c) * f^{-1}(d) \end{align}

- Therefore $f^{-1} : G \to G$ is an automorphism and $f^{-1} \in \mathrm{Aut} (G)$ is such that $f \circ f^{-1} = \epsilon$ and $f^{-1} \circ f = \epsilon$.

- Hence $(\mathrm{Aut}(G), \circ)$ is a subgroup of $(S_G, \circ)$, or more generally, $(\mathrm{Aut}(G), \circ)$ is a group. $\blacksquare$