Let $p$ be odd prime. $\Sigma=(p, \infty)$, $\mathbf{Q}_{\Sigma}$ is maximal Galois extension of $\mathbf{Q}$ unramified outside $\Sigma$. Let $G=\mathrm{Gal}(\mathbf{Q}_{\Sigma}/\mathbf{Q})$. $\Lambda=\mathbf{Z}_p[[X]]$. For any $n$, we define a character:
$\alpha: G \rightarrow (\Lambda/p^n)^{\times}$
Can I show that this character must factor through $(\mathbf{Z}/p^n)^{\times}$? I know use class field theory to show that if we replace $(\Lambda/p^n)^{\times}$ to $(\mathbf{Z}/p^n)^{\times}$ it will be satisfied. But it seems that I do not know about the structure about units of $(\Lambda/p^n)^{\times}$.
By the way, if it do not satisfy for arbitary $n$, does it satisfy for $n=1$?
The answer to both of your questions is "no". Notice that $\Lambda / p = \mathbf{F}_p[[X]]$. The cyclotomic character gives a surjection $G \to \mathbf{Z}_p^\times$, and it's easily checked that there's a unique homomorphism $\mathbf{Z}_p^\times \to \mathbf{F}_p[[X]]^\times$ sending the torsion subgroup of $\mathbf{Z}_p^\times$ to $1$ and sending $1 + p$ to $1 + X$. This map has infinite image, so it certainly does not factor through the finite group $\mathbf{F}_p^\times$.