91Ó°ÊÓ

We first need to define an action of \(G/A\) on \(A\). To do this, we take any element \(gA \in G/A\) and any \(a \in A\). Since \(A\) is normal in \(G\), we can consider the conjugate of \(a\) by \(g\). Define the action of \(gA\) as \((gA) * a = gag^{-1}\). We need to check that this defines a valid action. To do that, let's verify that this action satisfies the group action axioms: 1. Identity action: For this, we need to show that the identity of \(G/A\) acts as the identity on \(A\), which is \(eA * a = a\) for all \(a \in A\). Considering the action of \(eA * a = eae^{-1} = a\), it holds true. 2. Compatibility with group operation: For this, we need to show that \((g_1A * g_2A) * a = g_1A * (g_2A * a)\). Let \(g_1, g_2 \in G\). Then, we have \((g_1A * g_2A) * a = (g_1g_2A) * a = (g_1g_2) a (g_1g_2)^{-1}\). On the other hand, we have \(g_1A * (g_2A * a) = g_1A * (g_2ag_2^{-1}) = g_1(g_2ag_2^{-1})g_1^{-1} = (g_1g_2)ag_2^{-1}g_1^{-1}\). By associativity, these two expressions are equal. Now that we have shown that this operation defines an action of \(G/A\) on \(A\), we can proceed to the next step.

Now, we need to find a homomorphism between \(G/A\) and \(\operatorname{Aut}(A)\). For this purpose, define a map \(\phi: G/A \to \operatorname{Aut}(A)\) by \(\phi(gA) = \psi_{gA}\), where \(\psi_{gA}(a) = gag^{-1}\) for all \(a \in A\). We need to prove that \(\phi\) is a well-defined map and that it is a homomorphism.

First, to show that \(\phi\) is well-defined, note that if \(g_1A = g_2A\), then both elements are in the same coset of \(A\), which means \(g_1^{-1}g_2 \in A\) (or equivalently \(g_2^{-1}g_1 \in A\)). Thus, for any \(a \in A\), we have: \(g_1ag_1^{-1} = g_1(g_2^{-1}g_2)ag_1^{-1} = (g_1g_2^{-1})(g_2ag_2^{-1})g_2(g_1^{-1}g_2^{-1}) = (g_2^{-1}g_1)^{-1}(g_2ag_2^{-1})(g_2^{-1}g_1)\) As \(A\) is abelian, \((g_2^{-1}g_1)^{-1}(g_2ag_2^{-1})(g_2^{-1}g_1) = g_2ag_2^{-1}\). Thus, \(\psi_{g_1A} = \psi_{g_2A}\), and \(\phi\) is well-defined. Next, we will show that \(\phi\) is a homomorphism. Let \(g_1A, g_2A \in G/A\). We want to show that \(\phi(g_1A * g_2A) = \phi(g_1A) * \phi(g_2A)\). So we have: \(\phi(g_1A * g_2A) = \phi(g_1g_2A) = \psi_{g_1g_2A}\) \(\phi(g_1A) * \phi(g_2A) = \psi_{g_1A} * \psi_{g_2A} = \psi_{g_1A}(g_2ag_2^{-1})g_1^{-1} = g_1(g_2ag_2^{-1})g_1^{-1}\) Since \(A\) is abelian, this implies: \(\psi_{g_1g_2A}(a) = (g_1g_2)ag_2^{-1}g_1^{-1} = g_1(g_2ag_2^{-1})g_1^{-1} = \phi(g_1A) * \phi(g_2A)\) Therefore, we have shown that \(\phi\) is a homomorphism from \(G/A\) to \(\operatorname{Aut}(A)\), and our task is complete.