Theorem
Let \(G\) be a group and let \(a,b\in G\). Prove that the composition of the inner automorphisms corresponding to \(a\) and \(b\) satisfies
\[
\theta_a \circ \theta_b =\theta_{ab},
\]
where \(\theta_a\) and \(\theta_b\) denote the inner automorphisms corresponding to \(a\) and \(b\), respectively.
Proof
Step 1
Define the Inner Automorphisms
For any element \(g \in G\), the inner automorphism \(\theta_g\) is defined by:
\[
\theta_g(x) = g x g^{-1} \quad \text{for all } x \in G.
\]
Step 2
To compute the Composition \(\theta_a \circ \theta_b\)
For any \(x \in G\), we have:
\[
\begin{align}
(\theta_a \circ \theta_b)(x) &= \theta_a\big(\theta_b(x)\big) \nonumber\\
&= \theta_a\big(b x b^{-1}\big) \nonumber\\
&= a\,(b x b^{-1})\, a^{-1} \nonumber\\
&= (ab)\, x\, (b^{-1}a^{-1}) \nonumber\\
&= (ab)\, x\, (ab)^{-1}
\end{align}
\]
Step 3
Conclusion
Equation (1) shows that for all \(x \in G\):
\[
(\theta_a \circ \theta_b)(x) = \theta_{ab}(x).
\]
Hence, we conclude that:
\[
\theta_a \circ \theta_b = \theta_{ab}.
\]