Theorem
Prove that \(\operatorname{Inn}(G) \cong G/Z(G)\), where \(\operatorname{Inn}(G)\) is the group of inner automorphisms of \(G\) and \(Z(G)\) is the centre of \(G\).
Proof
Step 1
Define the Mapping
Define \(\phi: G \to \operatorname{Inn}(G)\) by
\[
\phi(g) = \theta_g, \quad \text{where } \theta_g(x) = g x g^{-1} \text{ for all } x \in G.
\]
Step 2
To show that \(\phi\) is a Homomorphism
For any \(g, h \in G\) and \(x \in G\), we have:
\[
\begin{align}
\phi(gh)(x) &= \theta_{gh}(x) = (gh)x(gh)^{-1} \nonumber\\
&= g\,(h x h^{-1})\,g^{-1} \nonumber\\
&= \theta_g\big(\theta_h(x)\big) \nonumber\\
&= (\theta_g \circ \theta_h)(x) \nonumber
\end{align}
\]
Thus, \(\phi(gh) = \theta_g \circ \theta_h = \phi(g) \phi(h)\), showing that \(\phi\) is a group homomorphism.
Step 3
To determine the Kernel of \(\phi\)
The kernel of \(\phi\) is given by:
\[
\ker(\phi) = \{g \in G \mid \theta_g(x) = x \text{ for all } x \in G\}.
\]
Since \(\theta_g(x) = g x g^{-1} = x\) for all \(x \in G\) if and only if \(gx = xg\) for all \(x \in G\), we have:
\[
\ker(\phi) = Z(G).
\]
Step 4
Apply the First Isomorphism Theorem
The First Isomorphism Theorem states that:
\[
G/\ker(\phi) \cong \operatorname{Im}(\phi).
\]
Since \(\operatorname{Im}(\phi) = \operatorname{Inn}(G)\) and \(\ker(\phi) = Z(G)\), it follows that:
\[
G/Z(G) \cong \operatorname{Inn}(G).
\]
