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