# Group Automorphisms

Recall that the groups $(G, \cdot)$ and $(H, *)$ are said to be isomorphic if there exists a bijection $f : G \to H$ such that for all $x, y \in G$ we have that $f(x \cdot y) = f(x) * f(y)$ and $f$ is said to be an isomorphism between$G$ and $H$.

We will now look at a special type of isomorphism which we define below.

Definition: Let $(G, *)$ be a group. An Automorphism of $G$ is a bijection $f : G \to G$ such that for all $x, y \in G$ we have that $f(x * y) = f(x) * f(y)$. |

*In other words, a group automorphism is an isomorphism from a group $G$ onto itself.*

Perhaps the simplest automorphism is the identity function. Let $\epsilon : G \to G$ be defined for all $x \in G$ to be:

(1)We've already seen that the identity function is a bijection. Let $x, y \in G$. Then:

(2)So for all $x, y \in G$ we have that $\epsilon (x * y) = \epsilon (x) * \epsilon (y)$, so $\epsilon$ is an automorphism of the group $(G, *)$.

We will now prove some a somewhat less trivial example of an automorphism.

Proposition 1: Let $(G, *)$ be an abelian group and let $\mu : G \to G$ be defined for all $x \in G$ by $\mu (x) = x^{-1}$. Then $\mu$ is an automorphism. |

**Proof:**We will first show that $\mu$ is a bijection. Let $x, y \in G$ and suppose that $\mu (x) = \mu (y)$. Then:

- Therefore $\mu$ is injective. Now let $y \in G$. Then if we let $x = y^{-1}$ we have that:

- So for each $y \in G$ there exists an $x = y^{-1} \in G$ such that $\mu (x) = y$, so $\mu$ is surjective. Since $\mu$ is both injective and surjective we have that $\mu$ by definition is a bijection.

- Now let $x, y \in G$. Then:

- Since $(G, *)$ is an abelian group, we see that $\mu (x * y) = \mu (x) * \mu(y)$ so $\mu$ is an automorphism. $\blacksquare$