Theorem
Let \( (G,\cdot) \) be a group and \( a\in G \). Define a mapping \( \theta_{a}:G\to G \) by
\( \theta_{a}(x)=a\cdot x \cdot a^{-1} \) for all \( x\in G \). Then:
- \( \theta_{a}\in \text{Aut}(G) \)
- \( \theta_{a}\circ\theta_{b}=\theta_{a\cdot b} \) for all \( a,b\in G \)
- \( (\theta_{a})^{-1}=\theta_{a^{-1}} \)
- \( \alpha\circ\theta_{a} \circ\alpha^{-1}=\theta_{\alpha(a)} \) for all \( \alpha\in \text{Aut}(G) \)
- \( G \) is abelian if and only if \( \theta_{a}=I_{G} \) for all \( a\in G \)
Proof
Part I
To prove \(\theta_a\) is an automorphism
We first show that \(\theta_a\) is a homomorphism. For any \(x,y\in G\):
\begin{align}
\theta_a(xy) &= a\cdot (xy) \cdot a^{-1} \nonumber\\
&= (a\cdot x \cdot a^{-1})(a\cdot y \cdot a^{-1}) \nonumber
\end{align}
Thus, \(\theta_a(xy)=\theta_a(x)\theta_a(y)\). Since the inverse map is given by \(\theta_{a^{-1}}\) (as will be shown in Part III), \(\theta_a\) is bijective. Therefore, \(\theta_a\in \text{Aut}(G)\).
Part II
To prove composition of conjugations
For any \(a,b\in G\) and for all \(x\in G\):
\begin{align}
(\theta_a\circ\theta_b)(x) &= \theta_a\big(\theta_b(x)\big) \nonumber\\
&= a\cdot (b\cdot x \cdot b^{-1}) \cdot a^{-1} \nonumber\\
&= (a\cdot b)\cdot x \cdot (a\cdot b)^{-1} \nonumber\\
&= \theta_{a\cdot b}(x) \nonumber
\end{align}
Thus, \(\theta_a\circ\theta_b=\theta_{a\cdot b}\) for all \(a,b\in G\).
Part III
To prove inverse of \(\theta_a\)
We claim that \((\theta_a)^{-1}=\theta_{a^{-1}}\). For any \(x\in G\):
\begin{align}
\theta_a\circ\theta_{a^{-1}}(x) &= \theta_a\big(a^{-1}\cdot x \cdot a\big) \nonumber\\
&= a\cdot (a^{-1}\cdot x \cdot a) \cdot a^{-1} \nonumber\\
&= x \nonumber\\
\end{align}
Similarly, \(\theta_{a^{-1}}\circ\theta_a(x)=x\). Therefore, \((\theta_a)^{-1}=\theta_{a^{-1}}\).
Part IV
To prove conjugation by an automorphism
Let \(\alpha\in \text{Aut}(G)\). For any \(x\in G\):
\begin{align}
(\alpha\circ\theta_a\circ\alpha^{-1})(x) &= \alpha\Big(\theta_a\big(\alpha^{-1}(x)\big)\Big) \nonumber\\
&= \alpha\Big(a\cdot \alpha^{-1}(x) \cdot a^{-1}\Big) \nonumber\\
&= \alpha(a)\cdot x\cdot \alpha(a)^{-1} \nonumber\\
&= \theta_{\alpha(a)}(x) \nonumber
\end{align}
Thus, \(\alpha\circ\theta_a\circ\alpha^{-1}=\theta_{\alpha(a)}\).
Part V
To prove characterization of abelian Groups
\(G\) is abelian if and only if for every \(a\in G\), \(\theta_a=I_G\). Suppose \(G\) is abelian. Then for any \(a,x\in G\):
\begin{align}
\theta_a(x) &= a\cdot x\cdot a^{-1} \nonumber\\
&= a\cdot a^{-1}\cdot x \nonumber\\
&= x \nonumber
\end{align}
Conversely, if \(\theta_a(x)=x\) for all \(a,x\in G\), then
\begin{align}
a\cdot x\cdot a^{-1} &= x \nonumber\\
\implies a\cdot x &= x\cdot a \nonumber
\end{align}
Thus, \(G\) is abelian.
