While studying algebra from Thomas Hungerford I have a question in proof of Theorem 7.11 on page 298.
It's image:
In last fifth line of proof I am not able to verify how the map $\sigma -> \bar {i} $ is an monomorphism.
How to prove that it's an homomorphism? The operation on left is a coposition and operation on right is multiplication.
So, $\sigma_{2} \sigma_{1} (u) $ is lhs of inequality but how to obtain rhs of this equality from it.
Kindly help.
If $\sigma(\zeta)=\zeta^i$ then I write $\sigma=\sigma_i$. Thus $$ \sigma_i(\sigma_j(\zeta))=\sigma_i(\zeta^j)=(\zeta^j)^i=\zeta^{ij} $$ hence $\sigma_i\sigma_j=\sigma_{ij}$. So we have a mapping $$ \Phi:Aut_KF\to \mathbb{Z}_n $$ with $\Phi(\sigma_i)=\overline{i}$. Then $$ \Phi(\sigma_i\sigma_j)=\Phi(\sigma_{ij})=\overline{ij}=\overline{i}\cdot \overline{j} $$