The <commutator subgroup> $[G,G]$ is characteristic: every automorphism sends commutators to commutators and therefore preserves the subgroup they generate. Thus an automorphism $A$ induces
$$
\overline A:G^{\mathrm{ab}}\longrightarrow G^{\mathrm{ab}},
\qquad g[G,G]\longmapsto A(g)[G,G].
$$
This is independent of the coset representative $g$. If $A'=c_h\circ A$ differs from $A$ by an <inner automorphism>, then
$$
A'(g)[G,G]=hA(g)h^{-1}[G,G]=A(g)[G,G]
$$
because the <abelianization> is abelian. Hence $q([A])=\overline A$ is independent of the representative of the outer class. Finally $\overline{AB}=\overline A\,\overline B$, so
$$
q:\operatorname{Out}(G)\longrightarrow\operatorname{Aut}(G^{\mathrm{ab}})
$$
is a well-defined group homomorphism.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 156 / 2 / b / Solution

{title2=$\operatorname{Inn}(G)$}
{wiki}

An inner automorphism is conjugation by an element of the group. The inner automorphisms form a normal subgroup $\operatorname{Inn}(G)\triangleleft\operatorname{Aut}(G)$.


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