Show that the kernel of the map \(G\to \aut(G)\) given by \(g\mapsto (h\mapsto gh\inverseof{g})\) is \(Z(G)\).
Solution
1. Write \(\varphi: G \to \aut(G)\) for the map \(\varphi(g) = c_g\), where \(c_g(h) = gh\inverseof{g}\).
2. \(g \in \ker \varphi\) if and only if \(g \in Z(G)\).
2.1. \(\ker \varphi = \ts{ g \in G \st c_g = \id_G }\), since the identity of \(\aut(G)\) is the identity automorphism.
2.2. \(c_g = \id_G\) says \(gh\inverseof{g} = h\) for every \(h \in G\).
2.3. Multiplying on the right by \(g\), this is equivalent to \(gh = hg\) for every \(h \in G\).
2.4. That is the defining condition for \(g \in Z(G)\).
3. Q.E.D.
Step 2 is the equality \(\ker \varphi = Z(G)\) of subsets, and both sides are subgroups of \(G\).