Let \(G\) be a group and \(A \trianglelefteq G\) be a normal abelian subgroup. Show that \(G/A\) acts on \(A\) by conjugation and construct a homomorphism \(\varphi: G/A \to \mathrm{Aut}(A)\).
Solution
For \(gA\in G/A\) and \(a\in A\), define \[ (gA)\cdot a=gag^{-1}. \]
1. The formula is well-defined and takes values in \(A\).
Normality of \(A\) gives \(gag^{-1}\in A\). Suppose \(gA=hA\). Then \(h^{-1}g\in A\), so write \(g=hb\) with \(b\in A\). Since \(A\) is abelian, \(bab^{-1}=a\), and therefore \[ gag^{-1}=h(bab^{-1})h^{-1}=hah^{-1}. \] Thus the value depends only on the coset \(gA\).
2. This formula defines a left action of \(G/A\) on \(A\).
The identity coset satisfies \[ A\cdot a=a. \] For \(g,h\in G\), \[ ((gA)(hA))\cdot a=(gh)a(gh)^{-1} =g(hah^{-1})g^{-1} =(gA)\cdot((hA)\cdot a). \] Hence the action axioms hold.
3. For each \(gA\in G/A\), conjugation by \(g\) restricts to an automorphismof \(A\).
Normality shows that conjugation by \(g\) maps \(A\) to itself. Its inverse on \(A\) is conjugation by \(g^{-1}\), so the restriction is an automorphism.
4. The map\[ \varphi:G/A\longrightarrow\operatorname{Aut}(A),\qquad \varphi(gA)(a)=gag^{-1}, \]is a group homomorphism.
3. For\(g,h\in G\) and \(a\in A\),\[ \varphi((gA)(hA))(a) =(gh)a(gh)^{-1} =\varphi(gA)(\varphi(hA)(a)). \]Thus \(\varphi((gA)(hA))=\varphi(gA)\varphi(hA)\).