Definition: Automorphism
An automorphism of a group \(G\) is an isomorphism from \(G\) onto itself. In other words, a map
\(\varphi: G \to G\) is an automorphism if it is a homomorphism, bijective, and its inverse is also a homomorphism.
Theorem
Let \((G,\cdot)\) be a group and let \(\alpha: G \to G\) be defined by \(\alpha(x) = x^{-1}\) for all \(x \in G\). Then \(\alpha\) is an automorphism if and only if \(G\) is abelian.
Proof
Step 1
Let \(\alpha\) is an Automorphism
Since \(\alpha\) is an automorphism, it is a homomorphism. Therefore, for all \(x, y \in G\),
\begin{align}
\alpha(xy) &= \alpha(x) \alpha(y) \tag{1}
\end{align}
By definition, \(\alpha(x) = x^{-1}\). Thus, equation (1) becomes:
\begin{align}
(xy)^{-1} &= x^{-1}y^{-1}. \tag{2}
\end{align}
However, from the general property of groups, we have:
\begin{align}
(xy)^{-1} &= y^{-1}x^{-1}. \tag{3}
\end{align}
Equating (2) and (3) gives:
\begin{align}
x^{-1}y^{-1} &= y^{-1}x^{-1}. \tag{4}
\end{align}
Taking inverses on both sides (noting that the inverse of an inverse returns the original element), we obtain:
\begin{align}
xy &= yx. \tag{5}
\end{align}
Hence, \(G\) is abelian.
Step 2
Let \(G\) is Abelian
Conversely, suppose \(G\) is abelian. Then for all \(x, y \in G\), we have:
\begin{align}
\alpha(xy) &= (xy)^{-1} \nonumber\\[5mm]
&= x^{-1}y^{-1} \quad \text{(since \(xy = yx\))} \nonumber\\[5mm]
&= \alpha(x)\alpha(y). \tag{6}
\end{align}
Thus, \(\alpha\) is a homomorphism. Also, since inversion is a bijective mapping on any group (with \(\alpha\) being its own inverse), \(\alpha\) is an automorphism.
