Group Automorphisms
Table of Contents

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:

\begin{align} \quad \epsilon (x) = x \end{align}

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

\begin{align} \quad \epsilon (x * y) = x * y \quad \mathrm{and} \quad \epsilon (x) * \epsilon (y) = x * y \end{align}

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:
\begin{align} \quad \mu (x) = \mu (y) \\ \quad x^{-1} = y^{-1} \\ \quad y * x^{-1} = y * y^{-1} \\ \quad y * x^{-1} = e \\ \quad [y * x^{-1}] * x = e * x \\ \quad y * [x^{-1} * x] = x \\ \quad y * e = x \\ \quad y = x \end{align}
  • Therefore $\mu$ is injective. Now let $y \in G$. Then if we let $x = y^{-1}$ we have that:
\begin{align} \quad y = (y^{-1})^{-1} = \mu(y^{-1}) = \mu(x) \end{align}
  • 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:
\begin{align} \quad \mu (x * y) = (x * y)^{-1} = y^{-1} * x^{-1} \quad \mathrm{and} \quad \mu (x) * \mu (y) = x^{-1} * y^{-1} \end{align}
  • Since $(G, *)$ is an abelian group, we see that $\mu (x * y) = \mu (x) * \mu(y)$ so $\mu$ is an automorphism. $\blacksquare$
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