Theorem
Let \(G\) be a group. Prove that \(|\operatorname{Inn}(G)| = 1\) if and only if \(G\) is commutative.
Proof
Step 1
Let us assume \(|\operatorname{Inn}(G)| = 1\)
Since \(|\operatorname{Inn}(G)| = 1\), the only inner automorphism of \(G\) is the identity map, denoted by \(\operatorname{I_{G}}\). Thus, for every \(g \in G\), we have:
\[
\theta_g(x) = gxg^{-1} = x \quad \text{for all } x \in G \nonumber
\]
Implies that:
\[
gx = xg \quad \text{for all } g,x \in G\nonumber
\]
Hence, \(G\) is commutative (abelian).
Step 2
Let us assume \(G\) is commutative
If \(G\) is abelian, then for every \(g, x \in G\) we have:
\[
\theta_g(x) = gxg^{-1} = xgg^{-1} = x\nonumber
\]
Therefore, each inner automorphism \(\theta_g\) is the identity map \(\operatorname{I_{G}}\). This shows that:
\[
\operatorname{Inn}(G) = \{\operatorname{I_{G}}\}\nonumber
\]
and consequently, \(|\operatorname{Inn}(G)| = 1\).